THE  NEW 

METAL  WORKER 


PATTERN  BOOK 


^^•KjrBiKn 


THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 


Arthur  H.  Memmler 
1874-1956 


THE   NEW 


METAL  WORKER 

PATTERN  BOOK 


A  TREATISE   ON   THE   PRINCIPLES   AND   PRACTICE   OF   PATTERN 
CUTTING   AS  APPLIED  TO  SHEET  METAL  WORK. 


BY    GEO.   W.    KITTREDGE. 


. NEW  YORK: 

DAVID  WILLIAMS  COMPANY,  232-238  WILLIAM  STREET. 

1901. 


THE  NEW  METAL  WORKER   PATTERN   BOOK, 
COPYRIGHTED,  1896,  BY  DAVID  WILLIAMS. 


THE  METAL  WORKER  PATTERN  BOOK, 

COPYRIGHTED,    l88l,    BY    DAVID   Wll.LlAMS. 


GIFT 


CONTENTS, 


PAGE. 
Introduction 

9 

CHAPTER    I. 
Terms  and  Definitions 

ALIMIABETIOAL    LlST    OF    TERMS     •  -  -  15 

CHAPTER    II. 
Drawing  Instruments  and  Materials  •  17 

CHAPTER    III. 
Linear  Drawing  - 

CHAPTER    IV. 

Geometrical  Problems 

CONSTRUCTION  OF  REGULAR  POLYGONS  -                                     43 

TIIK   KLLIVSE  •                  .         -       59 

TIIK   VOLUTE       -  -         -                  -         -           67 

CHAPTER    V. 

Principles  of  Pattern  Cutting  •       71 

PARALLEL  KOKMS  -                               72 

REGULAR  TAPERING   KOKMS  -       79 

IRREGULAR   KOKMS  -                           -                     86 

CHAPTER    VI. 

Pattern  Problems  -       96 

SECTION  1.  —  PARALLEL  FORMS  (MITEK  CUTTING)  »('» 

SECTION  2.  —  REGULAR  TAVKRING   FORMS  (FLARING  WORK)                                                                     240 

SECTION  3.  —  IRREGULAR  FORMS  (TKIAXGULATION)  306 


Index  of  Problems 


-    421 


766 


FOR  the  benefit  of  those  who  may  contemplate  making-  use  of  this  work,  wholly  or  in  part,  it  is 
well  to  lay  before  them  at  the  outset  a  general  statement  of  the  plan  upon  which  it  is  written, 
together  with  some  advice  for  the  use  and  study  of  the  same,  which  may  not  properly  belong 
under  any  of  the  several  headings  comprising  the  subject  matter.  A  glance  at  the  table  of  con- 
tents immediately  preceding  will  give  at  once  a  clear  idea  of  its  scope  and  arrangement.  From 
this  it  will  be  seen  that  the  first  five  chapters  are  theoretical  or  educational  in  their  nature, 
while  the  last  chapter  is  devoted  to  practical  work;  and  further,  that  the  book  does  not  presume 
upon  anv  previous  technical  knowledge  upon  the  part  of  the  beginner,  but  aims  to  place  before 
him  in  the  preliminary  chapters  all  that  is  necessary  to  a  thorough  understanding  of  the  work 
performed  in  the  last  chapter,  which  constitutes  the  bulk  of  the  book. 

A  very  important  feature  of  the  work  is  the  classification  of  the  problems.  The  forms  for 
which  patterns  mav  be  required  are  divided,  according  to  the  methods  employed  in  developing 
their  surfaces,  into  three  classes,  and  the  problems  relating  to  each  are  arranged  in  three  corre- 
sponding sections  of  the  last  chapter,  thus  bringing  near  together  those  in  which  principles  and 
methods  are  alike.  In  Chapter  V  (Principles  of  Pattern  Cutting)  this  classification  is  defined  and 
the  principles  governing  each  class  are  explained  and  illustrated  under  three  sub-headings  of  the 
chapter.  The  third  subdivision  treats  of  the  method  of  developing  the  surfaces  of  irregular 
forms  by  Triuiir/ulation,  a  subject  not  heretofore  systematically  treated  in  any  work  on  pattern 
cutting. 

A  chapter  on  drawing  (Chapter  III)  has  been  prepared  for  the  benefit  of  the  pattern  cutter 
especially  interested  in  cornice  wcrk,  and  though  he  may  not  intend  to  become  a  finished  archi- 
tectural draftsman,  this  chapter  will  render  him  valuable  assistance  in  reading  the  original  drawings 
received  from  architects,  from  which  he  is  required  in  many  cases  to  make  new  drawings  adapted 
to  his  own  peculiar  wants. 

The  New  Metal  Worker  Pattern  Book,  besides  being  a  systematic  treatise  on  the  principles  of 
pattern  cutting,  is  also  valuable  as  a  reference  book  of  pattern  problems  and  as  a  fund  of  information 
on  the  subject  treated,  to  be  drawn  from  at  convenience,  and  is  so  written  that  each  problem,  or  chapter 
of  descriptive  matter,  can  be  read  independently  of  the  others ;  so  that  the  student  whose  time  is 
limited  can  turn  to  any  portion  of  the  work  the  title  of  which  promises  the  information  sought, 
without  feeling  that  he  must  read  all  that  precedes  it.  The  relative  importance  of  the  chapters 
depends,  of  course,  upon  the  individual  reader,  and  will  be  determined  by  what  he  considers  his 
weakest  points.  However,  it  is  advisable  in  the  study  of  all  works  of  a  scientific  nature  to  begin 
at  the  beginning  and  take  everything  in  its  course.  If,  therefore,  the  study  of  this  work  can  be 
continued  progressively  from  the  first,  much  advantage  will  be  gained. 

The  statement   of  each  problem   in   prominent  type  appears  at  the  head  of  the  demonstration, 


iv  Tlie  Neiv  Metal    Worker  Pattern   Book. 

and  every  problem  is  numbered,  by  which  arrangement  the  problems  are  well  separated  from  each 
other  and  easily  found. 

While  each  demonstration  is  considered  complete  in  itself,  some  are  necessarily  carried  farther 
into  detail  than  others,  and  references  arc  made  from  one  problem  to  another,  pointing  out  similarity 
of  principle,  where  such  comparison  would  be  advantageous  to  one  who  is  looking  for  principles 
rather  than  for  individual  solutions. 

In  preparing  the  diagrams  used  to  illustrate  the  solutions  of  the  problems,  forms  have  been 
chosen  which  are  as  simple  in  outline  as  the  case  will  admit,  upon  the  supposition  that  the  reader 
will  be  able  to  make  the  application  of  the  method  described  in  connection  with  the  same  to  his  own 
especial  case,  which  may  embody  more  complicated  forms.  It  must  also  be  noted  that,  owing  to  the 
small  scale  to  which  the  drawings  in  this  work  are  necessarily  made,  extreme  accuracy  in  the 
operations  there  performed  is  impossible.  In  many  instances  the  length  of  the  spaces  used  in 
dividing  the  profiles  is  much  too  great  in  proportion  to  the  amount  of  curvature  to  insure  accuracy. 
Therefore  if  apparent  errors  in  measurements  or  results  are  found,  they  must  not  be  considered  the 
fault  of  the  system  taught.  If  such  errors  arc  discovered  the  student  is  recommended  to  reconstruct 
the  drawing  upon  his  own  drawing  board  in  accordance  with  the  demonstration  given  and  to  a  scale 
sufficiently  large  to  insure  accurate  results,  before  passing  judgment. 

In  the  preparation  of  this  book,  the  former  Metal  Worker  Pattern  Book  has  been  made  the  basis, 
to  a  certain  extent,  of  the  new  work.*  Such  problems  or  portions  of  the  former  work  as  were 
found  satisfactory  have  been  assigned  to  their  proper  places  in  the  new  work  without  change.  In 
the  case  of  most  of  the  problems,  however,  the  demonstrations  have  been  revised  and  the  drawings 
accompanying  them  have  been  amended  or  corrected  in  accordance  with  the  text,  and  in  many  cases 
entire  new  drawings  have  been  made.  To  these  have  been  added  a  large  number  of  new  problems 
based  upon  inquiries  and  solutions  that  have  appeared  in  the  columns  of  Tin-  MHul  W»rL-<'i-  since  the 
former  work  was  published.  Much  new  explanatory  matter  not  in  the  former  work  has  also  been 
added  in  the  preliminary  chapters,  prominent  among  which  are  Chapter  III,  and  the  principles  of 
Triangulation  in  Chapter  V. 

p]special  care  has  been  taken  in  the  composition  of  the  book  to  have  each  engraving  and  the 
text  referring  to  it  arranged,  as  far  as  possible,  on  the  same  page  or  upon  facing  pages,  so  as  to 
obviate  the  necessity  of  turning  the  leaf  in  making  references. 

A  great  advantage  is  gained  over  the  former  work  by  the  classification  and  numbering  of  the 
problems,  which,  in  connection  with  the  table  of  contents,  renders  any  desired  subject  or  problem 
easily  found. 

In  regard  to  the  system  of  reference  letters  employed  in  the  drawings,  it  should  be  said  that  the 
same  letter  has  been  used  so  far  as  possible  to  represent  any  given  point  in  the  several  views  or 
positions  in  which  it  may  occur,  the  superior  figure  or  exponent  being  changed  in  each  view.  To 
fully  comprehend  this  the  reader  must  carry  in  mind  the  concrete  idea  of  the  form  under  considera- 
tion, just  as  though  he  held  in  his  hand  a  perfect  completed  model  of  the  same,  which  lie  turned 
this  way  or  that  to  obtain  the  several  views  given.  Any  point,  therefore,  which  might  on  the  model 
be  marked  by  a  letter  A,  would  be  designated  in  one  of  the  views  as  A.  while  in  other  views  or 
places  where  it  might  appear  it  would  be  designated  as  A1,  A2,  etc.,  or  as  A',  A",  etc.  In  the 

*  Publisher's  Note.— The  author  of  this  book,  George  W.  Kittredge,  prepared  the  drawings  and  outlined  the 
demonstrations  of  all  but  a  few  of  the  less  important  problems  in  The  Metal  Worker  Pattern  Book,  which  was 
published  in  1881,  and  also  prepared  portions  of.  the  introductory  chapters  of  that  wcrk. 


Introduction* 

solution  of  problems  by  triangulation,  dotted  lines  arc  alternated  w-itli  solid  linos,  as  lines  of  meas- 
urement, merely  for  the  sake  of  distinction  and  to  facilitate  the  work. 

Occasions  arise  in  the  experience  of  every  pattern  cutter  wherein  some  portion  of  the  work 
before  him,  of  relatively  small  importance,  is  BO  situated  that  the  development  of  its  pattern  by  a 
strietly  accurate  method  would  involve  more  labor  and  time  than  would  be  justified  by  the  value 
of  the  part  wanted.  It  is  the  purpose  of  this  work  to  teach  the  principles  of  pattern  cutting,  leav- 
ing the  decision  of  such  questions  to  the  individual.  Nevertheless,  if  one  is  thoroughly  con- 
versant with  pattern  cutting  methods  and  familiar  with  pattern  shapes  it  may  be  possible  in  such 
cases  to  obtain  accurately  the  principal  points  of  a  required  pattern  and  to  complete  the  same  by  the 
eye  with  sullicient  accuracy  for  all  practical  purposes. 

As  intimated  above,  some  of  the  demonstrations  are' necessarily  made  more  explicit  than  others. 
In  the  longer  demonstrations  and  those  occurring  near  the  ends  of  the  Sections,  less  important  details 

x 

of  the  work  are  sometimes  omitted  and  certain  parts  of  the  operation  arc  only  hinted  at  or  are 
described  in  a  general  way,  upon  the  supposition  that  the  simpler  problems  in  which  the  demon- 
strations are  carried  further  into  detail  would  naturally  be  studied  first. 

Although  the  principles  of  pattern  cutting  here  set  forth  may  at  times  be  regarded  as  somewhat 
intricate,  it  is  believed  that  any  one  possessed  of  a  fair  degree  of  intelligence  and  application  can 
easily  master  them. 

Notwithstanding  the  great  care  which  has  been  used  in  the  preparation  of  this  work,  it  is 
possible  that  errors  mav  have  found  their  way  into  its  columns.  Should  errors  be  discovered  by  any 
of  its  readers,  information  of  such  will  be  gladly  received. 


CHAPTER    I. 


Pattern  cutting  as  applied  1<>  sheet-metal  work. 
by  its  verv  nature,  involves  ilic  application  <i{  geo- 
metrical principles.  Any  treatise  on  descriptive  geom- 
etry presents  in  a  general  way  all  the  principles  that 
enter  into  the  science  of  pattern  cutting.  To  those  who 
have  had  the  advantages  of  a  mathematical  education 
these  principles  are  well  known  and  l>v  such  their  ap- 
plication is  easily  made.  For  the  benefit  of  those, 
however,  who  have  not  had  such  advantages,  this  work 
purposes  to  make  specilic  application  of  those  princi- 
ples in  a  way  to  lie  readily  understood  bv  the  mechanic. 
While  throughout  the  work  the  use  of  an  unnecessarv 
number  of  technical  terms  and  words  not  in  common 
use  among  mechanics  will  be  carefullv  avoided,  it  must 
be  here  noted  that  precise  l«n</ii<i</f  in  describing  all 
geometrical  figures  and  operations  becomes  a  necessity, 
and  therefore  compels  the  employment  of  some  terms 
not  in  the  every  day  vocabulary  of  the  workshop,  which 
it  is  proper  to  delinc  ami  explain  at  the  outset.  As 
the  language  of  the  workshop  is  usuallv  far  from  ac- 
curate and  varies  with  the  locality,  every  student  of 
this  book  will  lind  it  greatly  to  his  advantage  to  give 
careful  attention  to  this  and  the  other  introductory 
chapters  for  the  purpose  of  increasing  and  improving 
his  vocabulary,  and  of  enabling  him  to  more  readily 
Comprehend  the  demonstrations  in  the  pages  following. 
The  list  of  terms  herein  defined  has  not  been  restricted 
to  the  barest  rccpiireniciits  of  the  book,  but  has  been 
made  to  include  nearly  all  the  terms  belonging  to  plain 
geometry,  and  such  architectural  terms  as  are  usually 
met  with  in  problems  relating  to  cornice  work.  The 
terms  are  arranged  first  logically,  in  classes,  after  which 
follows  an  alphabetical  list  by  which  an  v  definition  can 
be  readily  found. 

1.  Geometry  is  that  branch  of  mathematics  which 
treats  of  the  relations,  properties  and  measurements  of 
lines,  angles,  surfaces  and  solids. 

i'.  Sheet-Metal  Pattern   Cutting  is  founded   upon 

those  principles  of  geometry  which    relate  to   the  sur- 


faces of  solids,  and  may  be  more  accurately  described 
as  the  development  <>f  .<,///•/;<<•<*,  under  which  name  its 
principles  are  now  being  taught  to  a  great  extent  in 
schools  of  practical  instruction.  Articles  made  from 
sheet  metal  are  hollow,  being  only  shells,  and  must, 
therefore,  be  considered  in  the  process  of  pattern  cut- 
ting as  though  they  wen.'  the  coverings  or  casings 
stripped  from  solids  of  the  same  shape. 

3.  A  Point  is  that  which  has  place  or  position  with- 
out magnitude,  as  the  intersection  of  two  lines  or  the 
center  of  a  circle ;    it   is   usually  represented  to  the  eye 
by  u  small  dot. 

LINES. 

4.  A   Line  is  that  which   has  length   merely,  and 
may  be  straight  or  curved. 

5.  A  Straight  Line,  or,  as  it  is  sometimes  called,  a 
right  line,  is  the  shortest  line  that  can  be  drawn  between 
two  given   points.      Straight  lines  are  generally  desig- 
nated by  letters  or  figures  at  their  extremities,  as  A  B, 
Fig.  1. 

(i.    A  Curved  Line  is  one  which   changes  its  direc- 
tion at  every  point,  or  one  of  which  no  portion,  how- 

Fig.  1  —A  S  raight  Line. 

ever  small,  is  straight.  It  is  therefore  longer  than  a 
straight  line  connecting  the  same  points.  Curved  lines 


Fiy.  2. — Curved  Lines. 

are  designated  by  letters  or  figures  at  their  extrem- 
ities and  at  intermediate  points,  as  A  ~R  C  or  I)  K  I1', 
l-'iu'.  2. 


7.  Parallel  Lines  are  those  which  have  no  inclina- 
tion to  cadi  other,  being  everywhere  equidistant.  A 
B  and  A1  B'  in  Fig.  3  are  parallel  straight  lines,  and 


-B 

-B 


C'  D' 

Fig.  3.— Parallel  Lines. 

can  never  meet  though  produced  to  infinity.  G  D  and 
C'  D'  are  parallel  curved  lines,  being  arcs  of  circles 
which  have  a  common  center. 

8.  Horizontal  Lines  are  lines  parallel  to  the  hori- 
zon, or  level.  A  Horizontal  Line  in  a  drawing  is  indi- 
cated by  a  line  drawn  from  left  to  right  across  the 
paper,  as  A  B  in  Fig.  4. 


Horizontal 


Fig.  4. — Names  of  Lines  by  Direction. 

9.  Vertical   Lines  are    lines    parallel    to  a  plumb 
line  suspended  freely  in  a  still  atmosphere.     A  Verti- 
cal Line  in  a  drawing  is  represented  by  a  line  drawn 
up  and  down  the  paper,  or  at  right  angles  to  a  hori- 
zontal line,  as  E  C  in  Fig.  4. 

10.  Inclined  or  Oblique  Lines  occupy  an  interme- 
diate between  horizontal  and  vertical  lines,  as  C  D, 


Fig.  5.— Perpendicular  Lines. 

Fig.  4.  Two  lines  which  converge  toward  each  other 
and  which,  if  produced,  would  meet  or  intersect,  are 
said  to  incline  to  each  other. 


11.  Perpendicular  Lines. — Lines  are  perpendicular 

in  each  other  when  the  angles  on  either  side  ol  the 
point  of  meeting  are  equal.  Vertical  and  horizontal 
lines  are  always  perpendicular  to  each  other,  but  per- 
pendicular line.s  are  not  always  vertical  and  horizontal, 
but  may  be  at  any  inclination  to  the  horizon,  provided 
that  the  angles  on  either  side  of  the  point  of  intersec- 
tion are  equal.  In  Fig.  5,  C  F,  D  II  and  E  G  are 
said  to  be  perpendicular  to  A  B.  Also  in  Fig.  G, 


B  F 

Fig.  6. — Perpendicular  Lines. 

C  D  and  E  F  are  perpendicular  to  A  B.  Line?  per- 
pendicular to  the  same  line  are  parallel  to  each  other, 
as  C  F  and  D  H,  Fig.  5,  which  are  perpendicular 
to  A  B. 

12.  An  Angle  is  the  opening  between  two  straight 
linos  which  meet  one  another.  An  angle  is  commonly 
designated  by  three  letters,  the  letter  designating  the 
point  in  which  the  straight  lines  containing  the  angle 


B 


meet  being  between  the  other  two  letters,  as  the  angle 
BCD,  Fig.  4. 

13.  A  Right    Angle.— 'When   a  straight  line  meets 
another  straight  line  so  as  to  make  the  adjacent  angles 
equal  to  each  other,  each  angle  is  a  right  angle,  and 
the  straight  lines  are  said  to  be  perpendicular  to  each 
other.     (See  C  B  E  or  C  B  D,  Fig.  7.) 

14.  An  Acute  Angle  is  an  angle  less  than  a  right 
angle,  as  A  B  D  or  ABC,  Fig.  7. 

15.  An  Obtuse  Angle  is  an  angle  greater  than  a 
right  angle,  as  A  B  E,  Fig.  7. 


7'rrrnx    uml    Definitions. 


3 


STRAIGHT    SIDED    FIGURES. 

16.  A  Surface  is  that  which  has  length  and 
breadth  without  thickness. 

1  7.  A  Plane  is  a  surface  such  that  if  any  two  of 
its  points  be  joined  by  a  straight  line,  such  line  will  be 
wholly  in  the  surface.  Every  surface  which  is  not  a 
plane  surface,  or  composed  of  plane  surfaces,  is  a 
I'l-rri'd  surface. 

18.  A  Single  Curved  Surface  is  one  in  which  only 
certain  points  may  be  joined  by  straight  lines  which 


Fig.  8.— An  Equilateral  Triangle.    Fig.  9.— An  Isosceles  Triangle. 

B 
B 


Fig.  10.— A  Scalene  Triangle.      Fig.  11— Right- Angled  Triangles. 

shall  lie  wholly  in  its  surface.      The  rounded  surface  of 
a  cylinder  or  cone  is  a  single  curved  surface. 

19.  A  Double  Curved  Surface  is  one  in  which  no 
two  points  can  be  joined  by  a  straight  line  lying  wholly 
in  its  surface.     The  surface  of  a  sphere,  for  example, 
is  a  double  curved  surface. 

20.  A  Plane  Figure  is  a  portion  of  a  plane  termi- 
nated on  all  sides  by  lines  either  straight  or  curved. 

21.  A  Rectilinear  Figure  is  a  surface  bounded  by 
straight  lines.     (See  Figs.  8,  16,  21,  etc.) 

22.  Polygon  is  the   general    name  applied  to  all 
rectilinear  figures,  but  is  commonly  applied  to  those 
having  more  than- four  sides.      A  regular  polygon  is  one 
in  which  the  sides  are  equal. 

23.  A  Triangle  is  a  flat  surface  bounded  by  three 
straight  lines.     (Figs.  8,  9,  10,  11,  13,  etc.) 

24.  An  Equilateral  Triangle  is  one  in  which  the 
three  sides  are  equal.     (Fig.  8.) 


25.  An  Isosceles  Triangle  is  one  in  which  two  of 
the  sides  are  equal.      (Fig.  9.) 

26.  A  Scalene  Triangle  is  one  in  which  the  three 
sides  are  of  different  lengths.      (Fig.  10.) 

27.  A  Right-Angled  Triangle  is  one  in  which  one 
of  the  angles  is  a  right  angle.      (Fig.  11.) 

28.  An  Acute-Angled  Triangle  is  one  which  has  its 
three  angles  acute.     (Fig.  12.) 

29.  An  Obtuse-Angled  Triangle  is  one  which  has 
an  obtuse  angle.     (Fig  13.) 


Fig.  IS.— An  Acute-Angled 
Triangle. 


Fig.  IS.— An  Obtuse-Angled 
Triangle. 


Apex  orVcrt»X 


Base 


B 


Fig.  14.— Names  of  the  Sides  of  a       Fig.  IS.— Names  of  the  Parts 
Right- Angled  Triangle.  of  a  Triangle. 

30.  A  Hypothenuse  is  the  longest  side  in  a  right- 
angled  triangle,  or  the  side  opposite  the  right  angle. 
A  C,  Fig.  14. 

31.  The  Apex  of  a  triangle  is  its  upper  extremity, 
as  B,  Fig.  15.     It  is  also  called  vertex. 

32.  The  Base  of  a  triangle  is  the  line  at  the  bottom. 
B  C  and  A  C,  Figs.  14  and  15. 

33.  The  Sides  of  a  triangle  are  the  including  lines. 
A  C,  A  B  and  B  C,  Figs.  14  and  15. 

34.  The  Vertex  is  the  point  in  any  figure  opposite 
to  and  furthest  from  the  base.     The  vertex  of  an  angle 
is  the  point  in  which  the  sides  of  the  angle  meet.     B, 
Fig.  15. 

35.  The  Altitude  of  a  triangle  is  the  length  of  a 
perpendicular  let  fall  from  its  vertex  to  its  base,  as  B 
D,  Fig.  15, 

36.  A  Quadrilateral  figure  is  a  surface  bounded  by 
four  straight  lines.     There  are  three  kinds  of  Quadri- 


Tlie  \'.'»- 


M'  "•/.•<•/•   Pattern  Bool". 


laterals:   Tlie   Trapezium,  tin-  T  rape/old  and  the  J'ar- 


allelo<rram. 


37.  Tlio  Trapezium  has  no  t\vui>f  its  sides  parallel. 


(Fig.  n;.) 


38.   The  Trapezoid  has  only  two  of  its  sides  parallel. 

(Fig.  IT.) 

3i».  Tlu-  Parallelogram  has  its  opposite  sides  par- 
allel. There  are  four  varieties  of  parallelograms:  The 
Rhomboid,  the  Rhombus,  the  Rectangle  and  till- 
Square. 


411.    A   Heptagon    is  a  plane    figure  of  seven  sides. 
47.    An  Octagon   is  a  plane   figure  of   eight  sides. 


(Pig.  25.) 


4S.    A   Decagon  is   a   plane    figure    of    ten    sides. 


(Fig.  2<>.) 


49.  A  Dodecagon  is  a  plane  figure  of  twelve  sides. 
(Fig.  27.) 

5<i.  The  Perimeter  is  the  line  or  lines  bounding  any 
figure,  as  A  B  C  U  E,  Fig.  22. 

6 


Fig.  16.— A  Trapezium. 


Fig.  17.— A  Trapezoid. 


Fig.  22.— A  Pentagon.  Fiy.  23.— A  Hexagon.        Fig.  24.— A  Heptagon. 


Fig.  IS.— A  Rhomboid.  Fig.  19.— A  Rhombus  or  Lozenge. 


Fig.  15.— An  Octagon. 


Fig.  26— A  Decagon.  Fig. S7.—A  Dodecagon. 


Fig.  tO.— An  Equiangular  Par-    Fig.  il.—An  Equiangular 
allelogram  Called  a  Rectangle.       and  Equilateral  Parallel-        A~ 

ogram  Called  a  Square. 

40.  Tlie  Rhomboid  has    onlv  the    opposite    sides 
equal,  the  length   and   width   being    different  and   its 
angles  are  not  right  angles.      (Fig.  IS.) 

41.  The  FhombUS,  Lozenge  or  diamond  is  a  rhom- 
boid all  of  whose  sides  are  equal.      (Fig.   I'.'.) 

42.  The  Rectangle  is  a  parallelogram  all  of   whose 
angles  are  right  angles.      (Fig.  20.) 

43.  The  Square  is  an  equilateral  rectangle.     (Fig. 
21.) 

44.  A  Pentagon   is   a   plane    figure   of    live   sides. 
(Fig.  22.) 

45.  A   Hexagon   is  a   plane    figure    of    six    sides 
(Fig.  23.) 


Fig.  28.— Diagonals. 


'.  29.— A  Circle. 


51.  A  Diagonal  is  a  straight  line  joining  two  oppo- 
site angles  of  a  figure,  as  A  B  and  C  D,  Fig.  28. 

CIRCLES   AND  THEIR  PROPERTIES. 

52.  A  Circle  is  a  plane  figure  bounded  by  a  curved 

line,  everywhere  equidistant  from  its  center.  (Fig.  -'•'.) 
The  term  cirele  is  also  used  to  designate  the  boundary 
line.  (See  also  Circumference  ) 

53.  The  Circumference  of  a  circle  is  the  boundary 
line  of  the  figure.     (Fig.  29.) 

.~>4.  The  Center  of  a  circle  is  a  point  within  the 
circumference  equally  distant  from  every  point  in  its 
circumference,  as  A,  Fig.  29. 


Terms  and  Definitions. 


55.  The  Radius  of  a   circle   is   a  line  ilra\vn  from       dicnlar    to    the    radium   drawn    to    the    point    of    tan- 
the  center  to  any  point  in  the  circumference,  as  A   15.       gency.      Tims    K    I)    is   perpendicular  to  !•'  Hand  A  0 
Fig.  ait,  that  is,  half  the  diameter.     The  plural  of  radius       to  F  15. 

is  rinlii. 

56.  The  Diameter  of  a   circle  is  any  straight  line 
drawn  through  the  renter  to  opposite  points  of  the  cir- 
cumference, as  ('   I),    Fig.   L'!». 

57.  A   Semicircle   is   the   half  of  a  cirele,    and   is 


Dismeter  B 

Fig   SO.— A  Semicircle. 


Fiy.  SI — Stymints. 


bounded   liv  half   the   circumference  and    a   diameter 
(Fig.  3i».)  ' 

.V>.    A    Segment  of  a  circle  is  anv  part  of  its  sur- 
face    cut   oil'   l.v  a   straight  line,  as  A   K  15  and  C  F  D 
Fig-  31. 

.V.i.    An  Arc  of  a  circle  is  any  part  of   the  circum- 
ference, as  A  B  E  and  C  F  I),  Fig.  3:2. 


Fig.  fig — Arcs  and  Chords. 


Fiij.  .:.',.—  Sec/o  .1. 


(!<t.  A  Chord  is  a  straight  line  joining  the  extrem- 
ities of  an  arc.  as  A  E  and  C  D,  Fig.  3:2. 

61.  A   Sector  of  a  circle  is  the  space  included  be- 
tween   two   radii    and    the  arc  which  they  intercept,  as 
A  C  15  and  D  C  K.  Fig.  33,  and  B  A  C.   Fig.  :!4. 

62.  A   Quadrant  is  a  sector  whose  area  is  equal  to 
one-fourth  of  the  circle.     (B  A  C.  Fig.  34.)     The  two 
radii  bounding  a  quadrant  are  at  right  angles. 

ti:i.  A  Tangent  to  a  circle  or  other  curve  is  a 
straight  line  which  touches  it  at  only  one  point,  as  K  1) 
and  A  (_'.  Fig.  35.  Every  tangent  to  a  circle  is  perprn- 


Fig.  35. -Tangents. 

64.  Concentric  circles  are  those  which  are  described 
about  the  same  center.      (Fig.  36.) 

65.    Eccentric  circles  are   those  which  are  described 
about  different  centers.      (Fig.  37.) 

66.  Polygons  are  inscribed  in,  or  circumscribed  by, 


Fiy.  S6.  -  Concentric  Circles. 


Fig    -17  — Eccentric  Circles. 


circles  when  the  vertices  of  all  their  angles  are  in   the 
circumference.      (Fig.  3s.) 

67.  A  circle  is  inscribed  in  a  straight-sided  figure 
when  it  is  tangent  to  all  sides.  (Fig.  3!t.)  All  regular 
polygons  may  be  inscribed  in  circles,  and  circles  may 


Fig.  .»—  An  In'C'ibed  Tiiangle.     Fig   3!). -An  Inscribed  Circle. 

be  inscribed  in  the  polygons ;    hence   the   facility   with 
which  polygons  may  lie  constructed. 

•  >S.   A  Degree. — The  circumference  of  a  circle  is 
considered  as  divided  into  3t><»  equal  parts,  called  i/cr/rei-a 


Neiu  Metal    Worker  Pattern  ffook. 


(marked  °).  Each  degree  is  divided  into  60  minutes 
(marked  ') ;  and  each  minute  into  60  seconds  (marked  "). 
Thus  if  the  circle  be  large  or  small  the  number  of 
divisions  is  always  the  same,  a  degree  being  equal  to 


Fig.  40.  —  A  Circle  Divided  into  Degrees  for  Measuring  Angles. 


part  of  the  whole  circumference  ;  the  semicircle 
is  equal  to  180°  and  the  quadrant  to  90°.  The  radii 
drawn  from  the  center  of  a  circle  to  the  extremities  of 
a  quadrant  are  always  at  right  angles  with  each  other; 
a  right  angle  is  therefore  called  an  angle  of  90°  (A  E 
B,  Fig.  40).  If  a  right  angle  be  bisected  by  a  straight 
line,  it  divides  the  arc  of  the  quadrant  also  into  two 
equal  parts,  each  being  equal  to  one-eighth  of  the 
whole  circumference,  or  45°,  (A  E  F  and  FEE,  Fig. 
40)  ;  if  the  right  angle  were  divided  into  three  equal 
parts  by  straight  lines,  it  would  divide  the  arc  into 
three  equal  parts,  each  containing  30°  (AEG,  G  E  H, 
H  E  B,  Fig.  40).  Thus  the  degrees  of  the  circle  are 


Fig.  41.— Complement. 


Fig.  42.— Supplement. 


used  to  measure  angles,  therefore  by  an  angle  of  any 
number  of  degrees,  it  is  understood  that  if  a  circle  with 
any  length  of  radius  be  struck  with  one  foot  of  the 
compasses  in  its  vertex,  the  sides  of  the  angle  will 


intercept  a  portion  of  the  circle  equal  to  the  num- 
ber of  degrees  given.  Thus  the  angle  A  E  II,  Fig. 
40,  is  an  angle  of  60°.  In  the  measurement  of  angles 
by  the  circumference  of  the  circle,  and  in  the  various 
mathematical  calculations  based  thereon,  use  is  made 
of  certain  lines  known  as  circular  functions,  always 
liearing  a  fixed  relationship  to  the  radius  of  the  circle 
and  to  each  other,  which  gives  rise  to  a  number  of 
terms,  some  of  which,  at  least,  it  is  desirable  for  the 
pattern  cutter  to  understand. 

69.  The  Complement  of  an  arc  or  of  an  angle  is 
the  difference  between  that  arc  or  angle  and  a  quad- 
rant. In  Fig.  41,  A  D  B  is  the  complement  of  B  D  C, 
and  vice  versa. 

To.  The  Supplement  of  an  arc  or  of  an  angle  is  the 
difference  between  that  arc  or  angle  and  a  semicircle. 


Fig.  4$. — Diagram  Showing  the  Circular  Functions 
of  the  Arc  A  H  or  Angle  A  C  H. 

In  Fig.  42,  B  D  C  is  the  supplement  of  A  I)  B,  and 
vice  versa. 

71.  The  Sine  of  an  arc  is  a  straight  line  drawn 
from  one  extremity  perpendicular  to  a  radius  drawn  to 
the  other  extremity  of  the  arc.     (II  B,  Fig.  43.) 

72.  The  Co-Sine  of  an  arc  is  the  sine  of  the  com- 
plement of  that  arc.     H  K,  Fig.  4:1.  is  the  sine  of  the 
arc  A  H. 

73.  The  Tangent  Of  an  Arc  is  a  line  which  touches 
the  arc  at  one  extremity,  and  is  terminated  by  a  line 
passing  from  the  center  of  the  circle  through  the  other 
extremity  of  the  arc.     In  Fig.  43,  A  E  is  the  tangent 
of  A  H  or  of  the  angle  A  C  H. 

74.  The  Co-Tangent  of  an  arc  is  the  tangent  of 
the  complement.     Thus  F  G,  Fig.  43,  is  the  co-tan 
gent  of  the  arc  A  H. 


Terms  ami   Definitions, 


75.  Tlic  Secant  of  an  arc  is  a  straight  lino  drawn 
from  the  center  of  a  circle  through  one  extremity  of 
that  arc  and  prolonged  to  meet  a  tangent  to  the  other 
extremity  of  the  arc.  (K  0,  Fig  48.) 

70.  The  Co-Secant  of  an  arc  or  angle  is  the  secant 
of  the  complement  of  that  arc  or  angle,  as  V  C,  Fig. 
43. 

77.  The  Versed  Sine  of  an  arc  is  that  part  of  the 
radius  intercepted  between  the  sine  and  the  circumfer- 
ence.     (A  B,  Fig.  43.) 

78.  An  Ellipse  is  an  oval-shaped  curve  (Fig.  44), 


ix.  and  the  straight  lino  (CD)  the  directrix.  In  this 
lignre  anv  point,  as  N  or  M,  is  equally  distant  from 
F  ami  the  nearest  point  in  0  I>,  as  II  or  K.  (See  defi- 
nition 1  1 '.',.  i 

SO.  A  Hyperbola  (A  B.  Fig.  4ti)  is  a  curve  from  any 
point  in  which,  if  two  straight  lines  be  drawn  to  two 
fixed  points,  their  difference  shall  always  be  the  same. 
Thus,  the  difference  between  E  G  and  G  L  is  II  L. 
and  the  difference  between  E  F  and  F  L  is  B  L.  II  L 
and  B  L  are  equal.  The  two.  fixed  points,  E  and  L, 
are  called /oa.  (See  definition  113.) 


Fig.  44.— .In  KlHpse. 


Fig.  40.— A  Hyperbola. 


Fig.  45.— A  Parabola. 


Fig.  47.—Evolute  and  Involute. 


Fig.  4S.— A  Trian- 
gular Prism. 


from  any  point  in  which,  if  straight  lines  be  drawn  to 
two  fixed  points  within  the  curve,  their  sum  will  be 
always,  the  same.  These  two  points  are  called  foci  (F 
and  II).  The  line  A  B,  passing  through  the  foci,  is 
called  the  major  or  transverse  a.r/.s.  The  line  E  G,  per- 
pendicular to  the  middle  of  the  major  axis,  and  extend- 
ing from  one  side  of  the  figure  to  the  other,  is  called 
the  minor  or  conjugate  axis.  There  are  various  other 
definitions  of  the  ellipse  besides  the  one  given  here, 
dependent  upon  the  means  employed  for  drawing  it, 
which  will  be  fully  explained  at  the  proper  place 
among  the  problems.  (See  definition  113.) 

79.  A  Parabola  (A  B,  Fig.  45)  is  a  curve  in  which 
any  point  is  equally  distant  from  a  certain  fixed  point 
and  a  straight  line.  The  fixed  point  (F)  is  called  the 


81.  An  Evolute  is  a  circle  or  other  curve  from 
which  another  curve,  called  the  involute  or  evolutent,  is 
described  by  the  aid  of  a  thread  gradually  unwound 
from  it.      (Fig.  47.) 

82.  An  Involute  is  a  curve  traced  by  the  end  of  a 
string  wound  upon  another  curve  or  unwound  from  it. 
(Fig.  47.)     (See  also  Prob.  84,  Chapter  IV.) 

SOLIDS. 

83.  A  Solid  has  length,  breadth  and  thickness. 

84.  A  Prism  is  a  solid  of  which  the  ends  are  equal, 
similar  and  parallel  straight-sided  figures,  and  of  which 
the  other  faces  are  parallelograms. 

85.  A  Triangular  Prism  is  one  whose  bases  or  ends 
are  triangles.     (Fig.  48.) 


8 


n<-  \>->r   .\f't,il 


Pattern 


*•;.   A  Quadrangular  Prism  is  one  whose  bases  or  97.   A  Truncated  Cone  is  one  whose  apex  is  out  off 

ends  are  quadrilaterals.      (Fig.  49.)  l>y  a  plane  parallel  to  its  base.     (Fig.   .~>7.  )    This  1'iLnire 

87.  A  Pentagonal  Prism  is  one  whose  bases  or  ends  is  also  called  a  /;•;/.-•//////  of  a  cone.     A  pyramid  may  also 
are  pentagons.      (Fig.  50.)  be    ti:in<<-<ili<l.      (Sec    Figs.  t'>9    and    ~o    and    dctinition 

88.  A  Hexagonal  Prism  is  one  whose  bases  or  ends  112.) 

are  hexagons.  (Fig.  51.)  9S.  A  Pyramid  is  a  solid  having  a  straight-sided 

s'.t.  A  Cube  is  a  prism  of  which  all  the  faces  arc  base  and  triangular  .sides  terminating  in  one  point  or 

squares.  (Fiir.  ~>~2.  i  apex.  Pyramids  arc  distinguished  as  /rim n/u/rn-.  f/nn</- 

'.>(>.  A  Cylinde.",or  properly  sjieakingaCirCUlarCylin-  i-"ii>//t/ti>;  JM  iitnijnniil.  lir.rni/niiiil.  etc.,  according  as  the 

der,  is  a  round  solid  of  uniform  diameter,  of  which  the  base   has    three   sides,  four    sides,  live  sides,  six  sides. 

ends  or  bases  are  equal  and  parallel  circles.    (Fig..V!.j  etc.      (Figs.  .Vs;.  59  and  60.) 


Fig.  49.— A  Quad- 
rangular Prism. 


Fig.  50.— A  Pent-       Fig.  51.— A  Hex- 
agonal Prism.  agonal  Prism. 


Fig.  5S.—A  Cube.       Fig.  53.— A  Cylinder.    Fig  .14. — A  Cone. 


BOM 


Fig    55.—  A  Right 
Cone. 


Fig.  56  —An  Oblique 
or  Scalene  Cone. 


Fig.H7.-A  Tiun- 
cated  Cone. 


91.  Aii  Elliptical  Cylinder  is  one  whose  bases  are 
ellipses. 

92.  A  Right  Cylinder  is  one  whose  curved  surface 
is  perpendicular  to  its  bases. 

93.  An  Oblique  Cylinder  is  one  whose  curved  sur- 
face is  inclined  to  its  base. 

94.  A  Cone  is  a   round   solid  with  a  circle   for  its 
base,  and   tapering   uniformly    to   a    point  ac   the    top 
called  the  apex.     (Fig  54.) 

95.  A  Right  Cone  is  one  in  which  the  perpendicular 
let  fall   from  the  vertex  upon  the  base  passes  through 
the   center  of    the   base.      This   perpendicular  is  then 
called  the  axin  of  the  cone.     (Fig.  55.) 

'.»<;.   An  Obliqe  Cone  or  Scalene  Cone  is  one  in  which 
the  axis  is  inclined  to  the  plane  of  its  base.    (Fig.  5G.) 


Fig  .o.  -.1  Ti-ian-    Fig.  •'>!>.— .4  Quadran-    Fig.6().—An,Octag- 
gular  Pyramid.  gular  Pyramid.  onal  Pyramid. 


Fig.  61—  A   Right 
Pyramid. 


/• 


Fig.  r,2.—  Altitude        Fig    <;.!  —Altitude 
of  a  Cone.  of  a  Pyramid. 


\ 


\ 


fig.  r,4.-AUitude 
of  a  Priam. 


Fig.  r,5  —Altitude 
of  a  Cylinder. 


Fig.  66.— A  Sphere, 
or  Globe. 


99.  A  Right  Pyramid  is  one  whose  base  is  a  regular 
polygon,  and  in  which   the  perpendicular  let   fall  from 
the  apex   upon   the  base  passes  _ through  the  center  of 
the  base.     This  perpendicular  is  then  called   the  ".'/.-- 
of  the  pyramid.     (Fig.  61.) 

100.  The  Altitude  of    a    pyramid  or  cone  is  the 
length  of  the   perpendicular  let   fall   from  the  apex  to 
the  plane  of   the   base.      The   altitude   of   a   prism    or 
cvlinder  is  the  distance  between  its  two  bases  or  ends. 
and   is  measured    bv  a    line   drawn  from  a  point  in  one 
base    perpendicular   to   the   plane  of  the  other.      (Figs. 
:,<!.  il-2,  63,  64  and  65.) 

lul.  The  Slant  hight  of  a  pyramid  is  the  distance 
from  its  apex  to  the  middle  of  one  of  its  sides  at  the 
base.  The  xlimt  ///<////  of  a  cone  is  the  distance 


Terms  mul  Definitions. 


from    its  apex  to  any  point  in  the  eireumference  of   its 
base. 

l(>'2.  A  Sphere  or  Globe  is  a  solid  hounded  bv  a 
uniformly  curved  surface,  anv  point  of  which  is  equally 
distant  from  a  point  within  the  sphere  called  tin-  center. 
(Fig.  66.) 

103.  A   Polyhedron   is  a  solid   hounded  by  plane 
ligurcs.      There  are  live  regular  polyhedrons,  vi/,.  : 

104.  A  Tetrahedron  is  a  .solid  hounded   by   f<> in- 
equilateral   triangles.        It    is  one   form  of    triangular 
pyramid.      (.Fiji.   (>7.) 

105.  A    Hexahedron  is  a  solid    bounded    by  six 
squares.      The  common    name   for  this  solid   is 
which  see.      (Fig.  52.) 


io  each  other  as  to  appear  as  though  one  passes  through 

the  other.  The  intersection  of  their  surfaces  forms 
the  basis  of  the  greater  part  of  the  problems  of  Chap. 
VI. 

1 1  •>.  The  Frustum  of  a  Cone  or  Frustum  of  a  Pyra- 
mid is  that  portion  of  the  original  solid  which  remains 
after  the  apex  has  been  cut  away  upon  a  plane  parallel 
to  the  base..  (Figs.  .">".  <i!)  and  7u.)  When  the  cut- 
ting plane  is  oblique  to  the  base  of  the  solid  they  are 
spoken  of  as  nliliijur  frustums. 

113.  A  Conic  Section  is  a  curved  line  formed  by  the 
intersection  of  a  cone  and  a  plane*.  The  different 
conic  sections  are  the  triangle,  the  circle,  the  ellipse, 
the  parabola  and  the  hyperbola.  When  the  cutting 


Fig.  67.— A  Tetra- 
hedron. 


Fig.  CS.— An  Octa- 
hedron. 


Fig.  69.— Frustum 
of  a  Scalene  Cone. 


A  C 

Fig.  72.— A  Cone  Cut  by  a  Plane  Parallel 
to  One  of  Its  Sides. 


Fig.  73.— A  Cone  Cut  by  a  Plane  Which 
Makes  an  Angle  with  the  Base  Greater 
than  the  A  ngle  Formed  by  the  Side. 


Fig.  70.— Frustum 
of  a  Pyramid. 


Fig.  71. — .-1  Cone  Cut  by  a  Plane  Obliquely 
throu'jh  Its  Opposite  Sides. 


106.  The  Octahedron  is  a  solid  bounded  by  eight 
equilateral  triangles.  (Fig.  68.) 

lo".  The  Dodecahedron  is  a  solid  bounded  by 
twelve  pentagons. 

108.  The  Icosahedron  is  a  solid  bounded  by  twenty 
equilateral  triangles. 

109.  An  Axis  is  a  straight  line,  passing  through  a 
body  on   which   it   revolves,   or   may  be  supposed  to 
revolve.      (Figs.  55  and  61.) 

110.  By  the  Envelope  of  a  solid  is  meant  the  sur- 
face which  encases  or  surrounds  it,  as  the  envelope  of 
a  cone. 

111.  Intersection  Of  Solids  is  a  term  used  to  describe 
the  condition  of  solids  which  are  so  joined  and  fitted 


plane  passes  obliquely  through  its  opposite  sides  the 
resulting  figure  is  called  an  ellipse.  (Fig.  71.)  (An 
ellipse  is  also  an  oblique  section  through  a  cylinder.) 
When  a  cone  is  cut  by  a  plane  parallel  to  one  of  its 
sides,  the  resulting  figure  is  a. parabola.  Thus  in  Fig. 
72  the  cutting  plane  A  B  is  parallel  to  the  side  of 
the  cone  C  D.  See  definition  79.  When  the  cutting 
plane  makes  a  greater  angle  with  the  base  than  the  side 
of  the  cone  makes,  or  when  it  passes  vertically  through 
the  cone  to  one  side  of  the  axis,  the  resulting  figure  is 
a  hyperbola.  Thus  in  Fig.  73  the  angle  A  B  C  is  greater 
than  the  angle  ADR.  See  definition  80.  "  The  para- 
bola and  hyperbola  resemble  each  other,  both  beiny 
incomplete  figures,  with  arms  extending  indefinitely. 


10 


Tlie  New    Mrtitl    \\~,,, •/.•<;•    Pallrrn    Book. 


The  ellipse  is  a  complete  figure,  but  of  varying  pro- 
portions, as  the  cutting  plane  is  inclined  more  or  less. 
114.    Concave  means  hollowed  or  curved  inward. 


Fig.  74-— Sections  of  Curved  Surfaces. 

said  of  the  interior  of  an  arched  surface  or  curved  line 
in  opposition  to  convex.     (Fig.  74.) 

115.  A  Convex  surface  is  one  that  is  curved  out- 
ward, that  is  regularly  protuberant  or  bulging,  when 
viewed  from  without.     The  opposite  of  convex  is  con- 
cave.    (Fig.  74.) 

ARCHITECTURAL  TERMS. 

116.  The  term  Cornice  is  ordinarily  used  to  desig- 
nate any  molded  projection  or  collection  of  moldings 
which  finishes  or  crowns  the  part  to  which  it  is  affixed. 
The  term  in  this  sense  is  applicable  in  all  styles  of 
architecture.  In  classical  architecture,  however,  it  is  con- 
fined to  the  upper  division  of  the  entablature,  the  whole 


has  been  omitted.  The  names  of  parts  given  in  the 
illustration  are  such  as  are  generally  understood  l>r 
architects  and  cornice  makers.  The  cornice  of  clas- 
sical architecture  may  contain  simply  a  bed  mold, 
planceer  and  crown  mold,  or  it  may  contain,  in  addi- 
tion, a  dentil  course  or  a  modillion  course,  or  both. 

117.  The  Entablature  was  used  by  the  ancients  to 
finish  a  wall  or  colonnade  (more  especially  the  latter), 
and  consisted  of  three  parts,  the  cornice,  the  frieze  and 
the  architrave.     (Fig.  75.) 

118.  The  Architrave,  the    lower   division  of  the 
entablature,  was  in  reality  a  lintel  used  to  span  the 
space  between  the  columns,  but  its  form  was  main- 
tained when  used  above  a  wall.     In  modern  imitations 
of  the  antique  styles  the  molded  portion  is  frequently 
used  without  the  fascias,  in  which  case  it  is  commonly 
known  as  the  foot  mold.     (Fig.  75.)     The  term  arch- 
itrave is  also  used  to  designate  the  molding  and  fascias 
running  around  an  arch  or  a  window  opening. 

119.  The  Frieze,  the  middle  division  of  the  entab- 
lature, is  really  a  continuation  of  the  wall  surface  to 
add  hight  and  effect  to  the  building,  and  was  originally 
intended  for  the  display  of  symbols,  inscriptions,  orna- 
ments, &c.,  appropriate  to  the  use  of  the  building  of 
which  it  was   a  part.     It   is  sometimes  treated  very 
plainly  and  sometimes  receives  considerable  ornamen- 
tation, being  subdivided  into  panels  or  enriched  by 


ENTABLATURE  < 


CORNICE  < 

—       .....        .                     , 

J  CROWN 

,  PLANCEER   I 

MODILLION  BAND 

3 

__]'           IUvQ/^t)   V| 

££//       -k               (  MODILLION 
•  ==?i^^                    f     COURSE 

DENTIL  BAND 

T 

1 

1 

1 

1 

1 

1 

JDENTIL                                I  DENTIL 

STILE 

PANEL 

ARCHITRAVE  < 

STILE 

FASCIA 

FASCIA 

FASCIA 

Fig.  75.— The  Entablature  and  Its  Parts. 


of  which,  according  to  modern  ideas,  might  be  considered 
as  a  cornice.  The  distinction  is  shown  in  Fig.  75, 
which  shows  an  entablature  of  a  design  adaptable  to 
sheet  metal  construction,  and  in  which  all  enrichment 


scrolls,    etc.       The    terms  plain  frieze,   designating    a 
frieze  devoid  of  ornamentation,  md.  frieze-piece  or  //•/'<./ 
panel,  are  used  to  designate  one  of   the  parts  of  which  a 
frieze  is  constructed.     (Fig.  7/5.) 


Ti'/-iiis  ii/nl  Definitions. 


11 


120.  ArCi.  Tin1  curved  top  of  an  opening  in  a  wall. 
Tlio  arch  of  masonry  is  constructed  of  separate  blocks 
and  is  supported  only  at  the  extremities.  The  joint 
lines  between  the  bloeks  are  disposed  in  the  direc- 


plan,  designed  as  a  support  for  an  entablature.    It  con- 
sists of  three  parts  :   a  base,  a  shaft  and  a  capital. 

12-2.    An    Engaged    Column    is    a    column    placed 
against  the  face  of  a  wall  or  other  surface,  from  which 


Fig.  76.— A  Semicircular  Arch. 


Fig.  77.— A  Pointed  Arch. 


Fig.  80.— A  Pilaster. 


tion  of  radii  of  the  curve,  thus  enabling  the  arch  to 
support  the  weight  of  the  wall  above  the  opening. 
When  in  classical  designs  its  face  is  finished  with 
moldings  their  proper  profile  is  that  of  an  architrave. 
(Fig.  75.)  The  level  lines  at  which  the  curve  of  the 
arch  begins  are  called  the  springing  lines.  Sometimes 
the  lower  stones  of  the  arch  rise  vertically  a  short 
distance  from  the  supports  before  the  springing  lines 
are  reached,  in  which  case  the  arch  is  said  to  be  stilted. 


it  projects  one-half  or  more  than  one-half  its  diam- 
eter. 

123.  A  Pilaster  differs  from  a  column  in  that  it  is 
square  in  plan  instead  of  round  and  is  usually  engaged 
within  a  wall.     (Fig.  80.) 

124.  Pedastal.     A  structure  designed  to  support 
a  column,  statue,  vase  or  other  object.      It  is  by  some 
described    as   the   foot   of   a   column,  but  is,  properly 
speaking,    not   a   part   of    it.       It   consists   of    three 


Fig.  81. — An  Angular  Pediment. 


Fig.  78.— A  Moresque  Arch. 


Fig.  79.-A  Flat  Arch. 


Fig.  82.— A  Segmental  Pediment. 


The  stones  composing  the  arch  are  called  the  voussoirs, 
and  the  middle  or  top  stone  is  called  the  keystone.  The 
supports  below  the  ends  of  the  arch  are  called  imposts. 

Arches  are  usually  semicircular  (Fig.  76),  semi- 
elliptical,  segmental,  pointed  (Fig.  77)  or  Moresque  (horse- 
shoe) (Fig.  78)  in  shape,  according  to  the  style  of  archi- 
tecture with  which  they  are  used. 

The  top  of  an  opening  may  be  perfectly  level  and 
yet  composed  of  wedge-shaped  blocks  so  combined  as 
to  be  self-supporting,  in  which  case  it  is  called  a  flat 
arch.  (Fig.  79.) 

121.    A  Column  is  a  vertical  shaft  or  pillar  round  in 


parts,  a  base,  a  middle  portion  cubical  in  shape  called 
a  die  and  a  cap  or  cornice.  It  is  also  used  as  a  finish 
at  the  ends  of  a  balustrade  course. 


Fig.  SS.—A  Broken  Pediment. 

125.  A  Pediment  is  a  triangular  or  segmental  orna- 
mental facing  over  a  portico,  door,  window,  etc.  (Figs. 
81,  82  and  83. 


12 


TJie  Xeir 


Wor/,-> 


Ifook. 


126.  A  Broken  Pediment  is  one.  cither  in  the  f«rm 
of  a  gable  <>r  a  segment,  which  is  cni  away  in  its 
central  portion  for  the  purpose  of  ornamentation. 
(Fig.  83.) 

\-2~t.  A  Gable  is  the  vertical  triangular  end  of  a 
house  or  other  building,  from  the  cornice  or  eaves  to 

the   top. 

I'JS.  A  Lintel  Cornice  is  a  cornice  above  or  some- 
times including  a  lintel.  This  term  is  very  generally 
used  to  designate  the  cornice  used  above  the  lirst  story 
of  stores.  (Fig.  S4.) 

1 2!).  A  Deck  Cornice  or  Deck  Molding  is  the  cornice 


Fig.  S4.—A  Lintel  Cornice. 


Fig.  85.— A  Bracket. 

or  molding  used  to  finish  the  edge  of  a  flat  roof  where 
it  joins  a  steeper  portion. 

130.  A  Bracket,  as  used  in  sheet  metal  work,  is 
simplv  an  ornament  of  the  cornice.  Brackets  in  stone 
architecture  were  originally  used  as  supports  of  the 
parts  coming  above  them.  Hence  modern  architecture 
has  kept  up  that  idea  in  their  designs.  (Fig.  So.) 

1:51.  Modillions  are  also  cornice  ornaments,  and 
differ  from  brackets  onlv  in  general  shape.  (Fig.  86.) 
While  a  bracket  has  more'  depth  than  projection,  modil- 
lions  have  more  projection  than  depth. 

132.  A  Dentil  is  a  cornice  ornament  smaller  than 
a  modillion,  which  i,i  shape  usuallv  represents  a  solid 
with  plain  rectangular  face  and  sides.  Dentils  are 
never  used  singly,  but  in  courses,  the  spaces  between 
them  being  less  than  their  face  width.  (Fig.  76.) 


1:;:;.    A  Cortd  is  a  modified  form  of  bracket.    It  is 

used  to  terminate  the  lower  parts  of    window  caps,  and 
also  forms  the  support  for  arches,  etc.,  in  gothic forms. 


Fig.  S6.—A  Modillion. 


Fiij.  X7.—A  Head  Block. 


13-t.  A  Head  Block  or  Truss  is  a  large  terminal 
bracket  in  a  cornice,  projecting  suflicientl  v  to  receive 
all  tlu>  moldings  against  its  side,  thus  forming  a  finish 
to  the  end  of  the  cornice.  (Fig.  ST.) 

135.  A  Stop  Block  is  a  block-shaped  structure,  vari- 
ously ornamented,  which  is  placed  above  the  end 


Fig.  88.— A  Stop  Mock. 


Fig.  89.— A  Pinnacle. 


bracket  in  a'cornice,  and  which  projects  far  enough  to 
receive  against  its  side  the  various  moldings  occurring 
above  the  brackets,  forming  an  end  linish.  (Fig.  *V  i 
136,  A  Pinnacle  is  a  slender  turret  or  part  of  a 
building  elevated  above  the  main  building.  A  small 
spire.  (Fig.  s!».) 


Definitions. 


13 


137.    A  Finial  is  an   ornanie-nt.  variously  designed, 
placed    at    tlic    apex    of    a    pediment,    gable,    spire   or 

roof. 

13.s.   Capital. — The  upper  member  or  head  of  acol- 

iiniii  <>r  pilaster.    It  may  vary  in  character  according  to 

the  style  of  architecture  with  which  it  is  employed,  from 
a  few  simple  projecting  moldings  around  the  top  of  tin- 
column  to  an  elaborately  foliated  ornament.  The  lower- 

'A8ACUS 
IVOLUTE 


Fig.  00.— Capitals, 


most  mold  is  called  the  ni'i'l:  muld  and  the  uppermost 
member  sustaining  the  weight  of  the  lintel  or  arch 
above  is  called  the  ulnu-ii*.  (Fig.  90.) 

1:5!).  Panel. — A  sunken  compartment  having  molded 
cdu'cs  used  to  ornament  a  plane  surface',  as  a  frieze  ceil- 
ing, planceer  or  tvmpanum.  A  panel  may,  however, 
be  raised  instead  of  sunken. 

The  margin  or  space  between  the  sides  of  the  pane! 
and  the  edges  of  the  surface-  in  which  it  is  placed  is 
usuallv  made  equal  all  around  and  is  called  the  stil<j. 

140.  A  Volute  is  a  spiral  scroll  used  as  the  principal 
ornament  of  a  capital  and  is  placed  under  the  corners 
of  the  abacus.  For  inethe>il  of  drawing  the  volute  see 
Probs.  81  and  8i>,  Chap.  IV. 

14-1.  A  Molding  is  an  assemblage  of  forms  projevt- 
ing  bevoiid  the  wall,  column  or  surface-  to  which  it  is 
affixed.  (See  tirst  part  of  Chap.  V.) 

142.  Crown  Molding  is  the  term  applied  to  the  upper 
or  projecting  membe-r  of  a  cornice.      (Fig.  75.) 

143.  Planceer  or  PlanCher  is   the   ceiling  or  under 
side  of  the  projecting  part  of  a  cornice.      (Fig.  75.) 

144.  The  Bed  Moldings  of  a  cornice  are  those  mold- 
ings forming  the  lower  division  of   the  cornice  proper, 
and  which  are   made  up  of  the   bed   course,  modillion 
course  and  dentil  course.      (Fig.  75.) 

145.  The  Bed  Course  is  the  upper  division  of   the 
be-d  moldings,  the   part   with  which  the  bracket  heads 
and  modillion  heads  ordinarily  correspond,  and  against 
which  they  miter.      (Fig.  75.) 

14<i.  The  Modillion  Course  <>f  a  cornice  embraces 
the  modillions  and  all  the  moldings  which  are  imme- 
diately back  of  and  below  them.  The  plain  surface 
lying  back  of  or  between  the  modillions  is  called  in 


she-et  metal  work  the  ni'/'l.illinii    lauid.  anil   the  molding 
immediatelv   below  them  the    ///<////'////,/<  nmldiny.      (Fig. 

7:,.) 

147.  The  Dentil  Course  of  a  cornice  embraces  the 
dentils  and  all  tiie  moldings  to  which  the  dentils  are 
attached  as  ornaments,  comprising  the'/'/////  Imnd  and 
ill  niil  nKiii/iiiij.  (Fig.  75.) 

14S.  Foot  Molding  is  the  common  term  used  to 
elesignate  the  lower  molding  in  a  cornice.  It  is  fre- 
quently in  this  connection  used  in  the  sense  of  archi- 
trave. (Fig.  75.) 

149.  A  Bracket  Molding,  alsocalh-d  bracket  /,<«>l,  is 
the  molding  around   the   upper  part  of    a  bracket,  ami 
which  generally  members  with  the  bed  molding,  against 
which  it  finishes.      (Fig.  75.) 

150.  A  Gable  Molding  is  an  inclined  molding  which 
is  used  in  the  finish  of  a  gable. 

151.  A  Ridge  Molding  is  a  mohling  used  to  cap  or 
tinish  a  rielge.      It  is  also  culled  a  /•/'/<//•  fn/i/i///;/  or  sim- 
ply ridtjiii'j. 

152.  A  Hip  Molding  is  a  molding    used   to  protect 
anel  finish  the  hips  or  angles  of  a  n>of.      It  is  very  fre- 
quently  included  in  the  more  general  term  '•''</'//'//.</. 

153.  A  Fascia  is  a   plain  band  or  surface  below  a 
molding,  or,  in  other  words,  the  unornamente-el  face  of  a 
portion  of  a  cornice  or  architrave.      (Fig.  75.) 

154.  A  Fillet  is  a  narrow  plain  member  of  a  mold- 
ing   used    to    finish    or    separate    the    different    forms 
(a  a  a  a  Fig.  75  are  fillets.) 

155.  A   Drip   is   a  downward   projecting   member 
in  a  cornice  or    in  a   molding,  used  to  throw  the  water 
off  from  the  other  parts.      (Fig.  75.) 

15(i.  Sollit  is  the  term  applied  to  the  under  side 
of  a  projecting  molding,  cornice  or  arch. 

157.   A  Sink  is  a  depression  in  the  face  of  a  piece 


Fig.  91.— Stays. 


of  work  or  in  a  plain  surface.      (See  face  of  bracket 
Fig.  85,  side  of  modillion,  Fig.  sti.) 

158.  Incised  Work    is   a  style   of   ornamentation 

e-onsisting  of  line-  members  and  irregular  lines,  sunken 
or  cut  into  a  plain  surface.  (See-  side  of  bracket  Fig. 
85.) 

15!).    The  Stay  of  a  molding  is  its  shape  or  profile 
cut  in  sheet  metal.      (Fig.  lU.) 


u 


The  New  Metal    Worker  Pattern  Book. 


160.  Rake  Moldings  are  those  which  are  inclined, 
as  in  a  gable  or  pediment;  since  to  miter  a  rake  mold- 
ing with  a  level  return  under  certain  conditions  neces- 
sitates a  change  or  modification  of  profile   in  one  or 
the  other  of   the  moldings  to  rake  means  to  make  such 
change  of  profile. 

161.  A  Raked  Molding,  therefore,  is  a  term  describ- 


Fig.  92. — Elevation  of  a  Hoiise. 

ing  a  molding  of  which  the  profile  is  a  modification  of 
some  other  profile. 

162.  A  Raked  Profile  or  Raked  Stay  describes  the 
profile  or  stay  which  has  been  derived  from  another 
profile  or  stay,  by  certain  established  rules,  in  a  process 
like  that  of  mitering  a  horizontal  and  inclined  molding 
together. 

163.  The  Normal  Profile  or  Normal  Stay  is  the 
original  profile  or  stay  from  which  the  raked  profile  or 
stay  has  been  derived. 

B 


Fig.  9$.— Plan  of  a  House. 

164.  A  Flange  is  a  projecting  edge  by  which   a 
piece  is  strengthened  or  fastened  to  anything. 

165.  A  Hip  is  the  external  angle  formed  by  the 
meeting  of  two  sloping  sides  or  skirts  of  a  roof  which 
have  their  wall  plates  running  in  different  directions. 

DRAFTING    TERMS. 

166.  Projection  is  that  department  of  geometrical 
drawing   which  treats  of  the  drawing  of    elevations, 
plans,  sections  and  perspective  views.     There  are  four 


kinds  of  projections,  viz.  :  Orthographic,  Isometrical, 
Cavalier  and  Perspective.  Chapter  II I  is  devoted  to  an 
explanation  of  the  principles  of  orthographic  pro- 
jection. 


1 


Fig.  94. — Section  of  House  on  Line 
A  B  of  Plan  and  Elevation. 

167.  An  Elevation  is  a  geometrical  projection  of  a 
I  building  or  other  object  on  a  plane   perpendicular  to 

the  horizon.     (Fig.  92.) 

168.  A  Plan  is  the  representation  of  the  ]utrts  as 
they  would  appear  if  cut  by  a  horizontal  plane.     (Fig. 
93.) 

169.  A  Section  is  a  view  of  the  object  as  it  would 
appear  if  cut  in  two  by  a  given  vertical  or  horizontal 
plane.     (Fig.  94.)     In  the  one  case  the  resulting  view 
is  called  a  vertical  section,  and   in  the  other  a  horizontal 


fig.  05. — Perspective  View  of  House. 

section.     Oblique  sections  are  representations  of  objects 
cut  at  various  angles. 

170.  A  Perspective  is  a  representation  of  a  build- 
ing or  other  object  upon  a  plane  surface  as  it  would 
appear  if  viewed  from  a  particular  point.     (Fig.  9o.) 

171.  A  Detail  Drawing  or  Working  Drawing  is  a 
drawing,  commonly  full  size,  for  the  use  of  mechanics 
in  constructing  work. 

172.  A  Scale  Drawing  is  one  made  of  some  scale 
less  than  full  size. 


Ti.'i-inx  an'/.   Definitions. 


15 


173.  A  Miter  is  a  joint  in  a  molding,  or  between 
two  pieces  not  moldings,  at  any  angle. 

17i.  A  Butt  Miter  is  the  term  applied  to  the  cut 
made  upon  the  end  of  a  molding  to  lit  it  against 
another  molding  or  against  a  surface. 

175.  A  Gable  Miter  is    the    name    applied   to   the 
miter  either  at  the  peak  or  at  the  foot  of  the  moldings 
of  a  gable  or  pediment. 

176.  A  Rake  Miter  is  a   miter  between  two  mold- 
ings, one   of   which  lias   undergone   a   modification  of 
prolile  to  admit  of  the  joint  being  made. 

177.  Square    Miter    is    the    common   term    for   a 
joint  at  right  angles,  or  at  90°. 


17S.  An  Octagon  Miter  is  a  miter  joint  between 
two  sides  of  a  regular  octagon,  or  between  any  two 
pieces  at  an  angle  of  135°. 

1  7'.i.  An  Inside  Miter  indicates  a  joint  at  an  inte- 
rior or  re-entrant  angle. 

180.  An  Outside  Miter  is    a   joint  at  an  exterior 
angle. 

181.  The  Development  of  a  surface  is  the  process 
of  finding,  from  a  drawing  of  a  rounded  form,  a  shape 
or  pattern  upon  a  flat  surface  which,  when  cut  out  and 
bent  or  formed  as  indicated  by  that  drawing,  will  con- 
stitute its  envelope ;   or,  in  other  words,  the  stretching 
out  flat  of  a  surface  shown  by  a  drawing  to  be  curved. 


Alphabetical  List 


In   the    following   list   all  words   are  arranged  in  alphabetical  order,  the  figure  following  each  referring 
to  the  number  of  the  definition  in  the  list  preceding : 


Abacus 138 

Acute  Angle 14 

Altitude 35,  100 

Angle  12 

Angle,  Right 13 

Angle,  Acute 14 

Angle,  Obtuse 15 

Apex ' 31 

Arc  59 

Arch  120 

Architrave . 1 18 

Axis   109 

Base   32.   121,  124 

Bed  Course 145 

Bed  Moldings 144 

Bracket 130 

Bracket  Molding 149 

Broken  Pediment 126 

Butt  Miter 174 

Capital   138 

Center 54 

Chord  60 

Circle  52 

Circumference  53 

Circumscribed   66 

Column   121 

Complement   , 69 

Concave 114 

Concentric 64 

Cone  94 

Cone,  Oblique  or  Scalene..  96 


Cone,  Truncated 97 

Conic  Section 113 

Convex 115 

Corbel 133 

Cornice 116 

Co- Sine   72 

Co-Secant  76 

Co-Tangent 74 

Crown  Molding 142 

Cube  89 

Cylinder 90 

Cylinder,  Elliptical 91 

Cylinder,  Oblique 93 

Cylinder,  Right 92 

Decagon 48 

Deck  Cornice 129 

Degree  68 

Dentil 132 

Dentil   Course 147 

Detail  Drawing I7t 

Development 181 

Diagonal   51 

Diameter  56 

Die 124 

Dodecagon   49 

Dodecahedron   107 

Drip   155 

Elevation  167 

Ellipse 78 

Elliptical  Cylinder 91 

Engaged  Column 121 


Entablature 117 

Envelope   no 

Evolute 81 

Eccentric 65 

Fascia 153 

Fillet 154 

Finial   137 

Flange : 164 

Foot  Molding 148,  118 

Frieze 119 

Frustum 112 

Gable 127 

Gable  Miter 175 

Gable  Molding 150 

Geometry I 

Globe 102 

Head  Block 134 

Heptagon 46 

Hexagon  45 

Hexahedron  105 

Hip  165 

Hip  Molding 152 

Hyperbola 80 

Hypothenuse 30 

Icosahedron 108 

Impost    120 

Incised  Work 158 

Inscribed 66,  6- 

Inside  Miter 1/9 

Intersection  of  Solids in 

Involute 82 


Keystone' 120 

Line 4 

Line,  Curved 6 

Line,  Horizontal 8 

Line,  Inclined  or  Oblique..  10 

Lines,  Parallel 7 

Line,  Perpendicular n 

Line,  Straight 5 

Line,  Vertical 9 

Lintel   Cornice 128 

Lozenge 41 

Miter 173 

Modillion 131 

Modillion  Course 146 

Molding 141 

Neckmold  138 

Normal  Profile 163 

Oblique  Cone 96 

Oblique  Cylinder 93 

Obtuse  Angle 15 

Octagon 47 

Octagon  Miter 178 

Octahedron 106 

Outside  Miter 180 

Panel 139 

Parabola 79 

Parallelogram 39 

Pedestal 124 

Pediment 125 

Pediment.  Broken 126 

Pentagon 44 


16 


Tlie  Xc/c  Metal    \Vorker  Pattern  Book. 


Perimeter 50 

Perspective i/o 

Pilaster 123 

Pinnacle 136 

Plan 168 

Planccer 143 

Plane 17 

Plane  Figure 20 

Point 3 

Polygon 22 

Polyhedron  103 

Prism 84 

Prism,  Hexagonal 88 

Prism,  Pentagonal 87 

Prism,  Quadrangular 86 

Prism,  Triangular 85 

Projection  166 

Pyramid 98 

Pyramid.  Right 99 

Quadrant (12 

Quadrilateral   36 


Radius   55 

Rake 160 

Rake  Miter I/O 

Hake   Moldings 160 

Raked  Molding 161 

Raked  Profile 162 

Rectangle 4-' 

Rectilinear  Figure 21 

Rhomboid  40 

Rhombus 41 

Ridge   Molding 151 

Right   Angle 13 

Right  Cone 95 

Right  Cylinder 92 

Right  Line 5 

Right   Pyramid 99 

Scale  Drawing 172 

Scalene  Cone 96 

Scalene  Triangle 26 

Secant 75 

Section 169 


Sector 61 

Segment 58 

Semicircle  57 

Shaft 121 

Sides  of  a  Triangle 33 

Sink 157 

Sine 71 

Slant  Hight 101 

Soffit 156 

Solid 83 

Sphere   102 

Springing  Line 120 

Square   43 

Square  Miter 177 

Stay 159 

Stile 139 

Stilted 120 

Stop  Block 135 

Supplement  70 

Surface 16 

Surface,  Double  Curved..  .  19 


Surface,  Single  Curved....  18 

Tangent   63 

Tangent  of  an  Arc 73 

Tetrahedron  104 

Trapezium 37 

Trapezoid   38 

Triangle 23 

Triangle,  Aeute- Angled.  .  .  28 

'triangle,   Equilateral 24 

Triangle,  Isosceles 25 

Triangle,  Obtuse-Angled.  .  29 

Triangle,   Right- Angled.  ..  27 

Triangle.  Scalene 26 

Truncated   Cone 97 

Truncated    Pyramid 97 

Truss    134 

Versed  Sine 77 

Vertex   34 

Volute 140 

Vmissoir   120 

Working  Drawing 171 


CHAPTER   II. 


To  the  person  about  to  liegiu  ;i  new  occupation 
tin:  lirst  consideration  is,  what  tools  and  materials  does 
lie  need  ?  In  the  following  description  of  the  appli- 
ances, tools  and  materials  likely  to  be  of  service  to  the 
pattern  cutter  in  the  class  of  work  in  which  he  is  sup- 
posed to  he  the  most  interested,  the  description  is  limited 
to  articles  of  general  use.  Those  who  are  interested 
in  drawing  tools  and  materials  upon  a  broader  basis 
than  here  presented  are  referred  to  special  treatises  on 
drawing  and  to  the  catalogues  of  manufacturers  and 
dealers  in  drawing  materials  and  drawing  instruments. 

Drafting:  Tables.— A  drafting  table  suitable  for  a 
johbing  shop  .should  be  about  five  feet  in  length  and 
three  to  four  feet  in  width.  It  is  better  to  have  a 
table,  somewhat  too  large,  than  to  have  one  so  small 
that  it  is  frequently  inadequate  for  work  that  comes 
in.  In  hight  the  table  should  be  such  that  the  drafts- 
man, as  he  stands  up,  may  not  be  compelled  to  stoop 
ID  his  work.  While  for  some  reasons  it.  is  desirable 
that  the  table  should  be  fixed  upon  a  strong  frame  and 
legs,  for  convenience  such  tables  are  generally  made 
portable.  Two  horses  are  used  for  supports  and  a 
movable  drawing  board  for  the  top.  A  shallow  drawer 
is  hung  by  cleats  fastened  to  the  under  side,  and  is  ar- 
ranged for  pulling  either  way.  Sometimes  horizontal 
pieces  are  fastened  to  the  legs  of  the  horses,  and  a 
shelf  or  shelves  are  formed  by  laying  boards  upon 
them.  Fig.  96  shows  such  a  table  as  is  here  described. 
When  properly  made,  using  heavy  rather  t'han  light 
material,  such  a  table  is  quite  solid  and  substantial, 
and  when  not  in  use  can  be  packed  away  into  a  very 
small  space. 

For  cornice  makers'  use,  a  table  similar  in  con- 
struction to  the  one  described  and  illustrated  (Fig.  96) 
is  well  adapted.  Its  dimensions,  however,  consider- 
ing the  extremes  of  work  that  are  likely  to  arise, 
•ihould  be  twelve  to  fourteen  feet  in  length  by  about 
live  feet  in  breadth.  Three  horses  are  necessary,  aud 
two  drawers  may  be  suspended.  For  very  large  work, 


one  draftsman  or  pattern  cutter  will  require  the  whole 
table,  but  for  ordinary  work,  such  as  window  caps, 
cornices,  etc.,  two  men  can  work  at  it  without  interfer- 
ing w,ith  or  incommoding  each  other. 

Various  woods  may  be  used  for  drawing  tables, 
but  white  pine  is  the  cheapest  and  best  for  the  pur- 
pose. Inch  and  one-half  to  two-inch  stuff  will  be 
found  economical,  as  it  allows  for  frequent  redressing 
—made  necessary  by  pricking  in  the  process  of  pattern 
cutting.  Narrow  stuff,  tongued  and  grooved  together 
or  joined  by  glue,  is  preferable  to  wide  plank,  as  it  is 


Fig.  96.— Drafting  Table. 

less  liable  to  warp.  Eods  run  through  the  table  edge- 
ways, as  shown  in  Fig.  96,  are  desirable  for  drawing 
the  parts  together  and  holding  them  in  one  compact 
piece.  The  nut  and  washer  are  sunk  into  the  edge  of 
the  table,  a  socket  wrench  being  used  to  operate  them. 
A  drafting  table  should  be  an  accurate  rectangle 
— that  is,  every  corner  should  be  a  right  angle,  and 
the  opposite  sides  should  be  parallel.  The  edges 
should  be  exactly  straight  throughout  their  length. 
Methods  of  testing  drafting  tables  and  drawing  boards, . 
with  reference  to  these  points,  are  given  below.  The 
usual  way  of  adjusting  a  table  or  board  to  make  it  ac- 
curate is  to  plane  off  its  edges  as  required.  But  this 
is  a  task  less  simple  than  it  appeal's.  It  requires  the 
nicest  skill  and  accuracy  to  render  it  at  all  satisfactory. 


18 


Tlu:   \<-iv    Metal    Worker  Pattern  llouk. 


"When  it  is  remembered  that  110  matter  how  well  sea- 
soned the  lumber  employ ed  may  he  the  table  will  be 
affected  by  even  slight  changes  in  the  atmosphere,  it 
is  apparent  that  dressing  off  the  edges  with  a  plane, 
under  certain  circumstances,  might  be  constantly  re- 
quired. For  great  accuracy,  adjustable  metal  strips 
may  be  fastened  to  the  edges  of  the  table  in  such  a 
manner  that,  by  simply  turning  a  few  screws,  any 
variation  in  the  table  maybe  compensated.  This  ar- 
rangement maybe  accomplished  in  the  following  man- 
ner: The  edge  of  the  table  on  all  sides  is  cut  away  so 
as  to  allow  a  bar  of  steel,  say  one-eighth  or  one-six- 
teenth of  an  inch  thick  and  about  an  inch  wide,  to  lie 
in  the  cutting,  so  that  its  surface  is  even  with  the  face 
of  the  table,  with  its  outer  edge  projecting  somewhat 
beyond  the  edge  of  the  table.  Slotted  holes  are  made 
in  the  table  through  which  bolts  with  heads  counter- 
sunk into  the  metal  are  passed  for  holding  the  steel 
strips.  A  washer  and  nut  are  used  on  the  under  side 
of  the  table.  The  adjustment  required  is,  of  course, 
very  slight.  The  edge  of  the  metal  projecting  slightly, 
as  described,  is  well  adapted  for  receiving  the  head  of 
the  T-square,  rendering  the  use  of  that  instrument 
more  satisfactory  than  when  it  is  used  against  the  plane 
edge  of  the  table,  even  if  equally  accurate. 

Drawing;  Boards. — The  principal  difference  between 
a  drafting  table  and  a  drawing  board  is  in  the  size.  The 
same  general  requirements  in  point  of  accuracy,  etc., 
are  necessary  in  each.  Convenient  sizes  of  tables  for 
various  uses  Tiave  been  mentioned,  but  to  point  out 
sizes  of  boards  for  different  purposes  is  not  so  easv  a 
matter,  their  application  being  far  more  extended  and 
their  use  more  general.  A  drawing  board  may  be  made 
of  any  required  size,  from  the  smallest  for  which  such 
an  article  is  adapted  up  to  the  extreme  limit  con- 
sistent with  convenience  in  handling.  In  the  larger 
sixes  the  general  features  of  construction  noted  under 
drafting  tables  are  entirely  applicable,  save  that  thinner 
material  should  be  used  in  order  to  reduce  the  weight. 
In  small  sizes  there  is  a  choice  between  several  different 
modes  of  construction,  two  or  three  of  which  will  be 
described,  although  boards  of  almost,  any  required  con- 
struction can  be  purchased  ordinarily  of  dealers  in 
drawing  tools  and  materials  at  lower  prices  than  they 
can  be  made.  However,  it  is  very  convenient,  in  many 
cases,  to  have  boards  made-  to  order,  and  therefore  de- 
tailed descriptions  of  good  constructions  are  desirable. 
Any  carpenter  or  cabinet  maker  should  be  able  to  do 
the  work. 

In  Fig.  97  is  shown  a  very  common  form  of  draw- 


ing board,  consisting  of  a  pine  wood  top  with  hard- 
wood ledges.  The  ledges  are  put  on  by  means  of  a 
dovetail,  tapering  probably  one-half  inch  in  the  width 
of  the  board,  so  that  while  allowing  entire  freedom  for 


Fig.  97.— Drawing  Board,  with  Ledges. 

seasoning  there  is  no  danger  of  cracking  the  board, 
and  they  may  be  driven  tight  as  required.  "Where  it 
is  desirable  to  use  screws  in  the  ledges  they  arc  passed 
through  slotted  holes  furnished  with  a  metallic  bush- 
ing. 

In  Fig.  98  is  shown  a  still  simpler  form  of  board. 
which  is  adapted  only  for  the  smallest  sizes,     ilard- 


Fig.  98. — Drawing  Board,  with  Tongued  and  Grooved  Cleats. 

wood  strips  are  tongued  and  grooved  onto  the  ends  to 
prevent  warping,  as  shown  in  the  engraving.  By  using 
strips  of  wood  thicker  than  the  board,  keeping  their 
upper  surfaces  flush  with  the  surface  of  the  board,  it 
may  be  constructed  so  as  to  have  the  advantage  of 
ledges  on  the  under  side  equivalent  to  those  shown  in 
Fig.  97. 

Fig.  99  shows  a  construction  which,  while  being 


Fig.  99. — Bottom  View  of  Drawing  Board,  with  Grooved  Back  and 
Cleats  Attached  with  Slotted  Holes. 


somewhat  more  expensive  than  the  others,  is  un- 
doubtedly much  better.  Jt  is  made  of  strips  of  pine 
wood,  glued  together  to  make  the  required  width.  A 


19 


pair  of  hard-wood  cleats  is  screwed  to  the  hack,  the 
screws  passing  through  the  cleats  in  oblong  slots  with 
brass  bushings,  which  tit  closely  under  the  heads  and 
vet  allow  the  screws  to  move  freely  when  drawn  by  the 
shrinkage  of  the  board.  To  overcome  the  tendency  of 
the  surface  to  warp,  a  series  of  grooves  are  sunk  in  half 
the  thickness  of  the  board  over  the  entire  back.  To 
make  the  working  edges  perfectly  smooth,  allowing  an 
easy  movement  with  the  T-square,  a  strip  of  hard-wood 
is  let  into  the  end  of  the  board.  The  strip  is  afterward 
sawn  apart  at  about  every  inch,  to  admit  of  contraction. 
In  the  construction  of  such  boards  additional  advantage 
is  obtained  by  putting  the  heart  side  of  each  piece  of 
wood  to  the  surface. 

As  pattern  cutting  is  nothing  if  not  accurate,  it  is 
a  matter  of  the  utmost  importance  that  the  drawing 
board  or  table  should  be  perfectly  rectangular.  If  each 
angle  is  a  right  angle — if  its  opposite  sides  are  exactly 
parallel— the  T-square  may  be  used  at  will  from  any 
portion  of  it  with  satisfactory  results.  If  the  board  is 
accurate  the  drawing  will  be  accurate  If  the  board  is 
not  accurate  the  drawing  can  only  be  made  accurate  at 


Fig.  100.— Testing  the  Sides  of  a  Drawing  Board. 

the  cost  of  extra  trouble  and  care.  While  it  is  easy  to 
get  a  board  approximately  correct  by  ordinary  means, 
one  or  two  simple  tests  will  serve  to  point  out  inac- 
curacies for  correction  which  by  ordinary  means  would 
pass  unnoticed.  For  such  tests  a  T-square  and  an  or- 
dinary two-foot  steel  square  that  are  exactly  correct 
will  be  required. 

Having  made  the  opposite  sides  and  ends  of  the 
board  as  nearly  accurate  as  possible,  place  the  head  of 
the  T-square  against  one  side,  as  shown  in  Fig.  100, 
and  with  a  hard  pencil  sharpened  to  a  chisel  edge,  or 
with  the  blade  of  a  knife,  scribe  a  fine  line  across  the 
board.  Then  carrying  the  T-square  to  the  opposite 
side  of  the  board,  as  shown  by  the  dotted  lines,  bring 
the  edge  of  the  blade  to  the  line  just  scribed  and  see 
that  it  exactly  coincides  throughout  its  length  with  the 
line.  Eepeat  this  operation  at  frequent  intervals  along 
the  edges  of  the  board,  both  at  the  sides  and  ends. 
Remove  any  small  inaccuracies  on  the  edges  by  means 


of  a  file  or  line  sand  paper  folded  over  a  block  of 
wood.  Careful  work  in  this  manner  will  produce  very 
satisfactory  results. 

A  means  of  testing  a  board  with  reference  to  the 
accuracy  of  the  corners  is  shown  in  Fig.  101.  A  car- 
penter's try-square  or  an  ordinary  steel  square  used 


Fig.  101.— Testing  the  Corner  of  a  Drawing  Board. 

upon  the  corners  does  not  ordinarily  reach  far  enough 
in  either  direction  to  satisfactorily  determine  that  the 
adjacent  end  and  side  are  perpendicular  to  each  other; 
hence  it  is  desirable  to  obtain  some  kind  of  a  test  with 
reference  to  this  point  from  the  middle  portions  of  the 
edges.  With  the  head  of  the  T-square  placed  against 
one  side  of  the  board  draw  a  fine  line,  as  indicated  by 
the  dotted  line  in  the  engraving,  and  from  one  end  draw 
a  second  line  in  the  same  manner.  If  the  side  and  end 
are  at  right  angles  the  two  lines  will  coincide  with  the 
arms  of  a  square  when  placed  as  shown  in  the  engrav- 
ing. Eepeat  this  operation  for  each  of  the  corners. 
The  two  methods  above  described  for  testing  drawing 
boards,  especially  when  used  together,  cannot  fail  to 
enable  any  one  to  obtain  a  board  as  nearly  accurate  as 
it  is  possible  to  make  it.  Modifications  of  the  methods 
here  given,  and  based  upon  the  same  principles,  will 
suggest  themselves  to  any  one  who  will  give  the  mat- 
ter careful  thought. 

Straight-Edfes. — In  connection  with  every  set  of 
drawing   instruments  there   should    be   one    or   more 


o 


Fig.  102.—  Straight-Edge. 

straight-edges.  If  nothing  but  pencil  or  pen  lines  are 
to  be  made  upon  paper,  those  of  hard-wood  or  hard 
rubber  will  answer  very  well ;  but  if  lines  are  to  be 
drawn  upon  metal,  steel  is  the  only  satisfactory  ma- 
terial. The  length  of  the  straight-edge  must  be  de- 
termined by  the  work  to  be  done,  but  a  safe  rule  is  to 
have  it  somewhere  near  the  length  of  the  table  or 
board.  Of  course  this  is  out  of  the  question  in  cor- 
nice work,  where  tables  are  frequently  upward  of 


The.  New  Metal   Wurl,-<r   Pattern    fSuok. 


twelve  feet  in  length.  In  such  cases  the  size  of  the 
material  to  be  cut  determines  this  matter.  If  iron  96 
inches  long  is  used,  the  straight-edge,  for  convenience, 
should  not  be  less  than  8-J-  feet.  If  shorter  iron  is 
regularly  used,  a  shorter  straight-edge  will  answer.  In 
cornice  work,  two  and  even  three  different  lengths  are 
found  advantageous.  The  longest  might  be  as  just 
described;  a  second  might  be  about  four  feet  in 
length  and  made  proportionately  lighter,  while  the 
smallest  might  be  two  feet  and  still  lighter  than  the 
four-foot  size.  Instead  of  the  latter,  however,  the 
long  arm  of  the  common  steel  square  serves  a  good 
purpose. 

For  tinners'  \ise  in  general  jobbing  shops,  a  three- 
foot  straight-edge  in  many  cases,  and  a  four-foot  one 
in  a  few  instances,  will  be  found  very  convenient. 
Some  mechanics  desire  their  straight-edges  graduated, 
the  same  as  a  steel  square,  into  inches  and  fractions. 
There  is,  however,  no  special  advantage  in  this;  it 
adds  considerably  to  the  cost,  without  rendering  the 
tool  more  useful. 

A  hole  should  be  provided  in  one  end  of  the 
straight-edge  for  hanging  up.  It  should  always  be 
suspended  when  not  in  use,  as  in  that  position  it  is  not 
liable  to  receive  injury. 

It  is  almost  superfluous  to  add  that  straight-edges 
must  be  absolutely  accurate,  for  if  inaccurate  they 
would  belie  their  name.  A  simple  and  convenient 
method  of  testing  straight-edges  is  to  place  two  of 
them  together  by  their  edges,  or  a  single  one  against 
the  edge  of  a  square,  and  see  if  light  passes  between 
them.  If  no  space  is  to  be  observed  between  the 
edges  it  is  satisfactory  evidence  that  they  are  as  nearly 
straight  as  they  can  be  made  by  ordinary  appliances. 
In  addition  to  having  the  edges  straight  it  is  also 
necessary  to  have  the  two  sides  parallel. 

T-Squares. — With  this  instrument,  as  with  almost 
all  drawing  instruments,  there  is  the  choice  of  various 
qualities,  sizes  and  kinds,  and  selection  must  be  made 
with  reference  to  the  kind  of  work  that  is  to  be  perT 
formed.  Whatever  quality  may  be  chosen,  the  de- 
sirable features  of  a  T-square  are  strict  accuracy  in  ail 
respects,  and  a  thin,  flat  blade  that  will  lie  close  to  the 
paper.  For  most  purposes  a  fixed  head,  as  shown  in 
Fig.  103,  is  preferable.  For  drawings  in  which  a 
great  number  of  parallel  oblique  lines  are  required,  and 
particularly  where  a  small  size  J-square  can  be  used,  a 
swivel  head,  as  shown  in  Fig.  104,  is  sometimes  de- 
sirable. The  objectionable  feature  about  a  swivel  head 
is  the  difficulty  of  obtaining  positive  adjustment. 


\Vlien  made  in  the  ordinary  manner,  and  depending 
upon  the  friction  of  the  nut  of  a  small  bolt  for  holding 
the  head  in  place,  it  is  almost  impossible  to  obtain  a 
bearing  that  can  be  depended  upon  during  even  a 
simple  operation.  In  practice  it  is  found  to  be  far  less 
trouble  to  work  from  a  straight-edge — properly  placed 
across  the  board  and  weighted  down  or  otherwise  held 


Fig.  103.— l-Sifiiare  with  Fixed  Head. 

in    place — by   means   of   a   triangle   or   set-squaiv,    as 
greater  accuracy  is  thus  assured. 

In  point  of  materials,  probably  a  "[-square  having  a 
walnut  head  and  maple  blade  is  as  satisfactory  as  any. 
This  kind  is  the  cheapest  and  is  generally  considered 
the  best  for  practical  purposes.  A  good  article,  !>ut 
of  higher  price,  consists  of  a  walnut  head  with  a  Lard- 
wood  blade,  edged  with  some  oilier  kind  of  wood. 
Still  another  variety  has  a  mahogany  blade  edged  with 
ebony.  T-squares  constructed  with  cast-iron  head — 
open  work  finished  by  japanning — with  nickel-plated 
steel  blade,  are  also  to  be  had  from  dealers.  They  are 
also  made  with  a  hard  rubber  blade,  of  which  Fig.  104 
is  an  illustration.  The  liability  to  fracture,  however, 
by  dropping  necessitates  the  greatest  care  in  use: 
otherwise  hard  rubber  makes  a  very  desirable  article 
and  is  the  favorite  material  with  many  draftsmen. 

As  to  size,  T-squares  should  be  selected  with 
reference  to  the  use  to  be  made  of  them.  Generally, 
the  blade  should  be  a  very  little  less  in  length  than  the 
width  of  the  table  or  board  upon  which  it  is  to  be 


Fig.  W4.—t-Square  with  Fixed  and  Swivel  Bead. 

used.  Where  a  large  board  or  a  table  is  used  it  will 
be  found  economical  to  have  two  instruments  of  dif- 
ferent sixes. 

The  Ste.l  Square.— One  of  the  most  useful  tools  in 
connection  with  the  pattern  cutter's  outfit  is  an  ordinary 
steel  square.  The  divisions  upon  it  concern  him  much 
less  than  its  accuracy.  He  seldom  requires  other 
divisions  than  inches  and  eighths  of  an  inch;  therefore 
in  selection  the  principal  point  to  be  considered  is  that 


21 


of  aeeuracv.  The  finish,  bowever,  is  a  imittcr  not  to 
he  overlooked.  Since  a  nickel-plated  square  costs  hut 
a  trilling  advance  ii|nin  the  plain  article,  it  is' cheaper 
in  the  long'  run  to  have  the  plated  tool. 

A  convenient  method  of  testing  the  correctness  of 
the  outside  of  a  square,  and  one  which  can  be  used 
at  the  time  and  place  of  purchase,  is  illustrated  in 


Fig.  105,— Testing  the  Exterior  Angle  of  a  Steel  Square. 

Tig.  H>.">.  Two  squares  are  placed  against  each  other 
and  against  a  straight-edge,  or  against  the  arm  of  a 
third  square.  If  the  edges  touch  throughout,  the 
squares  mav  be  considered  correct. 

Having  procured  a  square  which  is  accurate  upon 
the  outside,  the  correctness  of  the  inside  of  another 
square  may  he  proven,  as  shown  in  Fig.  10t>.  Place 
mi"  square  within  the  other,  as  shown.  If  the  edges 


Fig.  Uifi.—Ttas  ing  the  Interior  Angle  of  a  Steel  Square. 

fit   together    tightly    and    uniformly    throughout,    the 
square  may  he  considered  entirely  satisfactory. 

An  accurate  square  is  especially  desirable,  as  it 
ail'ords  the  readiest  means  of  testing  the  "[-square  and 
the  drawing  table  and  beard,  as  elsewhere  described. 
The  greatest  care  should  be  given,  therefore,  to  the 
selection  of  a  square.  For  all  ordinary. purposes  the 


two-foot  size  is  most  desirable.  In  some  cases  the 
one-foot  sixe  is  better  suited.  Many  pattern  cutters 
on  cornice  work  like  to  have  both  sixes  at  their  com- 
mand, making  use  of  them  interchangeably,  according 
to  the  nature  of  the  work  to  he  done. 

Triangles,  or  Set  Squares.— In  the  selection  of  tri- 
angles, the  draftsman   has  the  choice  in  material  be- 


Fig.  107.— Open  Hard  Rubber  Triangle  or  Set  Sfauare,  45  x  43  x  90 
Degrees. 


tween  pear  wood ;  mahogany,  ebony  lined ;  hard  rub- 
ber ;  German  silver,  and  steel,  silver  or  nickel  plated. 
In  style  he  has  the  choice  between  open  work,  of  the 
form  shown  in  Fig.  107,  and  the  solid,  as  in  Fig. 
108.  In  shape,  the  two  kinds  which  are  adapted  to 
the  pattern  cutter's  use  are  shown  in  Figs.  107  and 
108,  the  latter  being  described  as  30,  60  and  90  de- 
grees, or  30  by  60  degrees,  and  the  former  as  45,  45 
and  90  degrees,  or  simply  45  degrees.  The  special 
uses  of  each  of  these  two  tools  are  shown  in  the 
chapter  on  Geometrical  Problems  (Chap.  IV).  In  size, 
the  pattern  cutter  requires  large  rather  than  small 


Fig.  lOS.—Hard  Wood  Triangle  or  Set  Square,  SO  x  GO  x  00  Degrees. 

ones.  If  he  can  have  two  sizes  of  each,  the  smaller 
should  measure  from  4  to  6  inches  on  the  side,  and 
the  larger  from  10  to  12  inches;  but  if  only  a  single 
size  is  to  be  had,  one  having  dimensions  intermediate 
to  those  named  will  be  found  the  most  serviceable. 

The  value  of  a  triangle,  for  whatever  purpose  used, 
depends  on  its  accuracy.  Particularly  is  this  to  be  said 
of  the  right  angle,  which  is  used  more  than  either  of 
the  others.  A  method  of  testing  the  accuracy  of  the 
right  angle  is  shown  in  Fig.  109.  Draw  the  line  A  B 


22 


Tin'  X»c  Mdal    Worker  Pattern  Boole. 


with  an  accurate  ruler  or  straight-edge.  Place  the 
right  angle  of  a  triangle  near  the  center  of  this  line,  as 
shown  by  D  C  B,  and  make  one  of  the  edges  coincide 
with  the  line,  and  then  draw  the  line  D  C  against  the 


•B— 


C 
Fig.  109.— Testing  the  Right  Angle  of  a  Triangle. 

other  edge.  'Turn  the  triangle  into  the  position  indi- 
cated by  D  C  A.  If  it  is  found  that  the  sides  agree 
with  A  C  and  C  D,  it  is  proof  that  the  angle  is  a  right 
angle  and  that  the  sides  are  straight. 

Besides  the  kinds  of  triangles  described  above,  a 
fair  article  can  be  made  by  the  mechanic  from  sheet 
zinc  or  of  heavy  tin.  Care  must,  however,  be  taken 
in  cutting  to  obtain  the  greatest  possible  accuracy.  For 
many  of  the  purposes  for  which  a  large  size  45  degree 


a  profile  line — is  called  spacers,  as  illustrated  and  de- 
scribed below. 

A  pair  of  compasses  consists  of  the  parts  shown 
in  Fig.  110,  being  the  instrument  proper  with  detach- 
able points,  and  extras  comprising  a  needle  point,  n 
pencil  point,  a  pen  and  a  lengthening  bar,  all  as  shown 
to  the  left.  In  selection,  care  should  be  given  to  the 
workmanship;  notice  whether  the  parts  lit  together 
neatly  and  without  lost  motion,  and  whether  the  joint 
works  tightly  and  yet  without  too  great  friction.  A 
good  German  silver  instrument,  although  quite  ex- 
pensive at  the  outset,  will  be  found  the  cheapest  in 
the  end.  A  pencil  point  of  the  kind  shown  in  our 
engraving  is  to  be  preferred  over  the  old  style  which 
clamps  a  common  pencil  to  the  leg.  The  latter  is  not 
nearly  so  convenient  and  is  far  less  accurate. 

Of  dividers  there  are  two  general  kinds,  the  plain 
dividers,  as  shown  in  Fig.  Ill,  and  the  hair-spring 
dividers,  as  shown  in  Fig.  112.  The  latter  differ  from 
the  former  simply  in  the  fact  of  having  a  fine  spring 
and  a  joint  in  one  leg,  the  movement  being  controlled 
by  the  screw  shown  at  the  right.  In  this  way,  after 


Fig.  110. — Compasses  with 
Interchangeable  Parts. 


Fig.  111.— Plain 
Dividers. 


Fig.    112.- Hair-Spring 
Dividers. 


Fig.  US.— Steel  Spring 
Spacers. 


triangle  would  be  used  the  steel  square  is  available, 
but  as  the  line  of  the  hypothenuse  is  lacking,  it  can- 
not be  considered  a  substitute. 

Compasses  and  Dividers.— The  term  compasses  is 
applied  to  those  tools,  of  various  sizes  and  descriptions, 
which  hold  a  pencil  and  pen  in  one  leg,  and  are  used  for 
drawing  circles,  while  dividers  are  those  tools  which, 
while  of  the  same  general  form  as  compasses,  have 
both  legs  ending  in  fixed  points,  and  are  used  for 
measuring  spaces.  A  special  form  of  dividers — used 
exclusively  for  setting  off  spaces,  as  in  the  divisions  of 


the  instrument  has  been  set  approximately  to  the  dis- 
tance desired,  the  adjustable  leg  is  moved,  by  means 
of  the  screw,  either  in  or  out,  as  may  be  required, 
thus  making  the  greatest  accuracy  of  spacing  possible. 
Both  instruments  are  found  desirable  in  an  ordinary 
set  of  tools.  The  plain  dividers  will  naturally  be  used 
for  larger  and  less  particular  work,  while  the  hair- 
spring dividers  will  be  used  in  the  finer  parts.  It  fre- 
quently happens  that  two  pairs  of  dividers,  set  to  dif- 
ferent spaces,  are  convenient  to  have  at  the  same  time. 
A  pair  of  spacers,  shown  in  Fig.  113,  is  almost 


Tool*    inn! 


23 


indispensable  in  a  pattern  cutter's  outfit.     He  will  find 

advantageous  use  for  this  tool,  oven  though  possess! IILT 
Imtli  pairs  of  dividers  described  above.  In  size  they 
are  made  less  than  that  of  the  dividers.  The  points 
should  be  needle-like  in  their  fineness,  and  should  lie 
capable  of  adjustment  to  within  a  very  small  distance 
of  each  other.  It  is  sometimes  desirable  to  divide  a 
given  profile  into  spaces  of  an  eighth  of  an  inch.  The 
spacers  should  be  capable  of  this,  as  well  as  adapted 
to  spaces  of  three-quarters  of  an  inch,  without  being 
too  loose.  As  will  be  seen  from  the  engraving,  this 
instrument  is  arranged  for  minute  variations  in  ad- 
justment. 

Beam  Compasses  and  Trammels — In  Fig.  114  is 

shown  a  set  of  beam  compasses,  together  with  a  portion 
of  the  wooden  rod  or  beam  on  which  they  are  used. 
The  latter,  as  will  be  seen  by  the  section  drawn  to  one 
side  (A),  is  in  the  shape  of  a  f.  This  form  has  con- 
siderable strength  and  rigiditv,  while  at  the  same  timr 
it  is  not  clumsy  or  heavy.  Beam  compasses  are  pro- 
vided with  extra  points  for  pencil  and  ink  work,  as 
shown.  While  the  general  adjustment  is  effected  by 
means  of  the  clamp  against  the  wood,  minute  varia- 
tions are  made  by  the  screw  shifting  one  of  the  points, 
as  shown.  This  instrument  is  quite  delicate  and  when 
in  good  order  is  very  accurate.  It  should  be  used  only 
for  line  work  on  paper  and  never  for  scribing  on  metal. 
A  coarser  instrument,  and  one  especially  designed 
for  use  upon  metal,  is  shown  in  Fig.  115  and  is  called 


Fig.  114.— Beam  Compasses, 

n  trammel.  It  is  to  be  remarked  in  this  connection 
that  the  name  trammel,  by  common  usage,  is  applied 
to  this  instrument  and  also  to  a  device  for  drawing 
ellipses,  which  will  be  found  described  at  another  place. 
There  are  various  forms  of  this  instrument,  all  being 
the  same  in  principle.  The  engraving  shows  a  form 


in  common  use.  A  heavier  stick  is  used  with  it  than 
with  the  beam  compasses,  and  no  other  adjustment  is 
provided  than  that  which  is  all'orded  by  clamping 
against  the  stick.  In  the  illustration  a  carrier  at  the 


Fig.  115. — Trammel. 

side  is  shown  in  which  a  pencil  may  be  placed.  Some 
trammels  are  arranged  in  such  a  manner  that  either  of 
the  points  may  be  detached  and  a  pencil  substituted. 

A  trammel,  by  careful  management,  can  be  made 
to  describe  very  accurate  curves,  and  hence  can  be  \ised 
in  place  of  the  beam  compasses  in  many  instances. 
For  all  coarse  work  it  is  to  be  preferred  to  the  beam 
compasses.  It  is  useful  for  all  short  sweeps  upon 
sheets  of  metal,  but  for  curves  of  a  very  long  radius  a 
strip  of  sheet  iron  or  a  piece  of  wire  will  be  found  of 
more  practical  service  than  even  this  tool. 

The  length  of  rods  for  both  beam  compasses  and 
trammels,  up  to  certain  limits,  is  determined  by  the 
nature  of  the  work  to  be  done.  The  extreme  length 
is  determined  by  the  strength  and  rigidity  of  the  rod 
itself.  It  is  usually  convenient  to  have  two  rods  for 
each  instrument,  one  about  3$  or  i  feet  in  length  and 
the  other  considerably  longer — as  long  as  the  strength 
of  material  will  admit.  In  the  case  of  the  trammel, 
by  means  of  a  simple  clamping  device,  or,  in  lieu  cf 
better,  by  use  of  common  wrapping  twine,  the  rods 
may  be  spliced  when  unusual  length  is  required;  but 
a  strip  of  sheet  iron  or  a  piece  of  fine  wire  forms 
a  better  radius,  under  such  circumstances,  than  the 
rod. 


24: 


The  New  Metal   Worker  Pattern  book. 


The  Protractor  is  an  instrument  for  laying  down 
and  measuring  angles  upon  paper.  The  instrument 
consists  of  a  semicircle  of  thin  metal  or  horn,  as  rep- 
resented in  Fig.  116,  the  circumference  of  which  is 
•divided  into  180  equal  parts  or  degrees.  The  princi- 
ples upon  which  the  protractor  is  constructed  and  used 
are  clearly  explained  in  the  chapter  on  Terms  and  Defi- 
nitions (Def.  68  "  Degree  ").  The  methods  of  employing 
it  in  the  construction  of  geometrical  figures  are  shown 


Fig.  116. — Semicircular  Protractor. 

in  Chapter  IV  among  the  problems.  For  purposes  of 
accuracy,  a  large  protractor  is  to  be  preferred  to  a 
small  size,  because  in  the  former  fractions  of  a  degree 
are  indicated. 

While  a  number  of  geometrical  problems  are  con- 
veniently solved  by  the  use  of  this  instrument,  it  is 
not  one  that  is  specially  adapted  to  the  pattern  cutter's 
use.  All  the  problems  which  are  solved  by  it  can  be 
worked  out  by  other  accurate  and  expeditious  methods, 
which,  in  most  cases,  are  preferable.  It  is  one  of  the 
instruments,  however,  included  in  almost  every 
case  of  instruments  sold,  and  the  student  will  find 
it  advantageous  to  become  thoroughly  familiar  with 
it,  whether  in  practice  he  employs  it 
or  not. 

Besides  the  semicircular  form  of  the  .,,, 

protractor  shown,  corresponding  lines  and  Ln  iilinn 

divisions  to  those  upon  it  are  sometimes 
put  upon  some  of  the  varieties  of  scales 
in  use,  as  shown  in  Fig.  120. 

Scales. — Many  of  the  drawings  from  which  the 
pattern  cutter  works — that  is,  from  which  he  gets  di- 
mensions, etc., — are  what  are  called  scale  drawings, 
being  some  specified  fraction  of  the  full  size  of  the 
object  represented.  Architects'  elevations  and  floor 
plans  are  very  generally  made  either  -J-  or  -J  inch 
to  the  foot,  or,  in  other  words,  -5*3-  or  -fa  full  size. 
Scale  details  are  also  employed  quite  extensively  by 
architects,  scales  in  very  common  use  for  the  purpose 


being  H  inrhes  to  the  foot  and  3  inches  to  the  fool, 
or,  in  other  words,  ^  and  -J-  full  size  respectively.  It 
is  essential  that  the  pattern  cutter  should  be  familiar 
with  the  various  scales  in  common  use,  that  lie  may 
be  able  to  work  from  any  of  them  on  demand.  Sev- 
eral of  the  scales  are  easily  read  by  means  af  the  com- 
mon rule,  as,  for  example,  3  inches  to  the  foot,  in 
which  each  quarter  inch  on  the  rule  becomes  one  inch 
of  the  scale;  also,  1-J-  inches  to  the  foot,  in  which 
each  eighth  of  an  inch  on  the  rule  becomes  an  inch  of 
the  scale;  and,  likewise,  f  inch  to  the  foot,  in  which 
each  sixteenth  of  an  inch  on  the  rule  becomes  an  inch 
of  the  scale.  However,  other  scales  besides  these  are 
occasionally  required,  which  are  not  easily  read  from 
the  common  rule,  and  sometimes  special  scales  are 
used,  which  are  not  shown  on  the  instruments,  espe- 
cially calculated  for  the  purpose.  Accordingly,  it  is 
sometimes  necessary  for  the  pattern  cutter  to  construct 
his  own  scale. 

The  method  of  constructing  a  scale  of  1  inch  to 
the  foot  is  illustrated  in  Fig.  117,  in  which  the  di- 
visions are  made  by  feet,  inches  and  half  inches.  In 
constructing  such  scales,  it  is  usual  to  set  off  the  di- 
visions representing  feet  in  one  direction  (say  to  the 
right)  from  a  point  marked  0,  while  the  divisions  for 
inches  and  fractions  thereof  are  set  off  the  opposite 
way  (or  to  the  left  from  0)  as  shown  in  the  illustra- 
tion. In  using  the  scale,  measurements  are  made  by 
placing  one  point  of  the  dividers  at  the  number  of 
feet  required ;  the  other  point  can  then  be  moved  to 
the  other  side  of  the  0  to  the  required  number  of 
inches,  thus  embracing  the  entire  number  of  feet  and 
inches  between  the  points  of  the  dividers. 

Besides  scales  of  the  kind  just  described,  which 


Inches 


Feet 


Fig.  117.— Plain  Scale  (1  inch  to  the  Foot  ) 


are  termed  plain  divided  scales,  there  are  in  common 
use  what  aBe  known  as  diagonal  scales,  an  illustration 
of  one  of  which  is  shown  in  Fig.  118.  The  scale  rep- 
resented is  that  of  1-j-  inches  to  the  foot.  The  left- 
hand  unit  of  division  has  been  divided  by  means  of 
the  vertical  lines  into  12  equal  parts,  representing 
inches.  In  width  the  scale  is  divided  into  8  equal 
parts  by  means  of  the  parallel  lines  running  its  entire 
length.  Next  the  diagonal  lines  are  drawn,  as  shown. 


J)ra>ci>ig   7W<  and  Materials, 


Bv  a  moment's  inspection  it  will  bo  seen  that,  by 
means  of  these  diagonal  lilies,  one-eighth  of  an  inch 
and  multiples  thereof  are  shown  on  the  several  hori- 
zontal lines.  A  distanee  equal  to  the  space  from  A  to 
11.  as  marked  on  the  scale,  is  read  (first  at  the  right  for 
feet)  '2  feet  (then  to  the  left  for  inches  by  means  of 


Aflat  scale  is  also  manufactured  in  both  boxwood 
and  ivory.  Fewer  scales  or  divisions  can  be  put  upon 
it  than  upon  the  triangular  scale,  yet  for  certain  pur- 
poses it  is  to  be  preferred  to  the  latter.  There  are 
less  divisions  to  perplex  the  eye  in  hunting  out  just 
what  is  required,  and  accordingly,  there  is  less  lia- 


Fent 


Fig.  US.—  Diagonal  Scale 


inches  to  the  foot). 


the  vertical  lines  ligured  both  at  top  and  bottom)  6 
inches  (and  last,  by  means  of  the  diagonal  line,  figured 
at  the  end  of  the  scale,  for  fractions)  and  three-eighths. 
The  top  and  bottom  lines  of  the  scale  measure  feet  and 
inches  only.  The  other  horizontal  lines  measure  feet, 
inches  and  fractions  of  an  inch,  each  horizontal  line 
having  its  own  particular  fraction,  as  shown.  Such 
scales  are  frequently  quite  useful,  as  greater  accuracy 
is  obtained  and,  as  the  reader  will  see,  may  be  con- 
structed by  any  one  to  any  unit  of  measurement,  and 
divided  by  the  number  of  horizontal  lines  into  any  de- 
sired fractious. 

A  scale  in  common  use,  and  known  as  the  tri- 
angular scale,  is  shown  in  Fig.  110.  The  shape  of 
this  scale,  which  is  indicated  by  the  name,  and  which 


bility  to  error  in  its  use.  However,  the  limited  num- 
ber of  scales  which  it  contains  greatly  restricts  its 
usefulness. 

Fig.  120  shows  another  form  of  the  flat  scale,  in 
quite  common  use  in  the  past,  but  now  virtually  dis- 
carded in  favor  of  more  convenient  dimensions  and 
shapes.  This  scale  combines  with  the  various  divi- 
sions of  an  inch  the  divisions  of  the  protractor,  as 
shown  around  the  margin.  The  fact  that  the  divisions 
of  an  inch  for  purposes  of  a  scale  are  located  in  the 
middle  of  the  instrument,  away  from  the  edge,  which 
makes  it  necessary  to  take  off  all  measurement  with 
the  dividers,  renders  the  article  awkward  for  use,  and 


Fig.  119  — Triangular  Boxwood  Scale. 

is  also  shown  in  the  cut,  presents  three  sides  for  divi- 
sion. By  dividing  each  of  these  through  the  center 
lengthways  by  a  groove,  as  shown,  six  spaces  for 
divisions  are  obtained,  and  by  running  the  scales  in 
pairs — that  is,  taking  two  scales,  one  of  which  is  twice 
the  size  of  the  other,  and  commencing  with  the  unit 
at  opposite  ends — the  number  of  scales  which  may  be 
put  upon  one  of  these  instruments  is  increased  to 
twelve.  This  article,  which  may  be  had  in  either 
boxwood,  ivory  or  plated  metal,  and  of  6,  12,  18  or 
24  inches  in  length,  is  probably  the  most  desirable  for 
general  use  of  any  sold. 


s 

2 

j 

\       ' 

S 

\ 

i 

1 

f'      , 

/ 

/    /    / 

'       - 

.,•• 

f 

•^ 

-^ 

~i'f^ 

-^ 

\    .. 

0 

i  •> 

-V 

„- 

'•',r 

^ 

^Q//^J 

2 

f< 

t; 

0 

1 

2 

| 

-1 

B 

-I 

7 

e 

^ 

10 

1  1 

12J    13 

M 

n 

16 

*' 

a 

I-.-, 

0 

1 

2 

3 

4 

B 

f 

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0 

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0 

1 

2 

I 

4 

,'_ 

. 

1lN 

(j 

1 

2 

; 

=  s 

Fig.  ISO. — Flat  Scale  with  Divisions  of  the  Protractor  on  the  Margins. 


the  arrangement  of  the  divisions  of  the  circle,  on  the 
margins,  is  less  satisfactory  for  use  than  the  circular 
protractor. 

Lead  Pencils. — Various  qualities  of  pencils  are  sold, 
some  at  much  lower  prices  than  others,  but,  all  things 
considered,  in  this  as  in  other  case's,  the  best  arc  the 
cheapest.  The  leading  brands  arc  made  in  two  grades 
or  qualities.  The  ordinary  grades  employ  numbers. 
1,  2,  3,  etc.,  to  indicate  hardness  of  lead,  No.  1  being 
the  softest,  and  No.  5  being  the  hardest  in  common 
use.  A  finer  grade  of  pencils,  known  as  poligrades,  is 
marked  by  letters,  commencing  at  the  softest  with  B  B, 


T/ie  New  Metal    Worker  Pattern,  Book. 


and  ending  at  the  hardest  with  H  H  II  II  H  H,  while 
other  makes  of  pencils  are  marked  by  systems  peculiar 
to  their v manufacturer.  The  draftsman  has  the  choice 
of  round  or  hexagon  shape  in  all  except  the  finest 
grades,  the  latter  being  made  exclusively  hexagon. 
Whatever  kind  of  pencil  the  draftsman  or  mechanic 
uses,  he  will  require  different  numbers  for  different 
I  mi-poses.  For  working  drawings,  full-sized  details, 
etc.,  on  manila  paper,  a  No.  3  (or  F)  is  quite  satis- 
factory. Some  like  a  little  harder  lead,  and  therefore 
prefer  a  No.  -i  (or  II).  For  lettering  and  writing  in 
connection  with  drawings  upon  manila  or  ordinary 
detail  paper,  a  No.  2  (II  B)  is  usually  chosen.  For 
fine  lines,  as  in  developing  a  miter,  in  which  the  great- 
est possible  accuracy  is  required,  a  No.  5  is  very  gen- 
e rally  used,  although  many  pattern  cutters  prefer  the 
liner  grade  for  this  purpose  and  use  a  H  II  II  H  H. 

The  quality  and  accuracy  of  drawings  depend,  in 
a  considerable  measure,  upon  the  manner  in  which 
pencils  arc  sharpened.  A  pencil  used  for  making  fine 


Fig.  121.—  Two  Views  of  Pencil  Sharpened  to  a  Chisel  Point. 

straight  lines,  as,  for  instance,  in  the  various  opera- 
tions of  pattern  cutting,  should  be  sharpened  to  a  chisel 
point,  as  illustrated  in  Fig.  121.  Pencils  for  general 
work  away  from  the  edges  of  the  T-square,  triangle, 
etc.,  should  be  sharpened  to  a  round  point,  as  shown 
in  Fig.  122.  It  facilitates  work  and  it  is  quite  eco- 
nomical to  have  several  pencils  at  command,  sharpened 
in  different  ways  for  different  purposes.  Where  for 
any  reason  only  one  pencil  of  a  kind  can  be  had,  both 
ends  may  be  sharpened,  one  to  a  chisel  point  and  the 
other  to  a  round  point. 

For  keeping  a  good  point  upon  a  pencil,  a  piece 
of  fine  sand  paper  or  emery  paper,  glued  upon  a  piece 
of  wood,  will  be  found  very  serviceable.  A  flat  file, 
mill-saw  cut,  is  also  useful  for  the  same  purpose. 
Sharpen  the  pencil  with  a  knife,  so  far  as  the  wood 
part  is  concerned,  and  then  shape  the  lead  as  required 
upon  the  file  or  sand  paper. 

Drawing  Pens.— Although  most  of  the  pattern 
cutter's  work  is  done  with  the  pencil,  there  occasion- 
ally arise  circumstances  under  which  the  use  of  ink  is 
desirable.  Tracings  of  parts  of  drawings  are  frequently 


required  which  can  be  better  made  with  ink  than  with 
pencil. 

The  drawing  pen  or  ruling  pen,  as  illustrated  in 
Fig.  123,  is  used  for  drawing  straight  lines.  The 
drawing  pen,  whether  as  a  separate  instrument  or  as 
an  attachment  to  compasses  or  beam  compasses  for 
drawing  curved  lines,  consists  of  two  blades  with  steel 
points,  fixed  to  a  handle.  The  blades  are  so  curved 
that  a  sufficient  cavity  is  left  between  them  for  ink 
when  the  points  meet  close  together  or  nearly  so.  The 
space  between  the  points  is  regulated  by  means  of  the 
screw  shown  in  the  engraving,  so  as  to  draw  lines  of 
any  required  thickness.  One  of  the  blades  is  provided 
with  a  joint,  so  that,  by  taking  out  the  screw,  the 
blades  may  be  completely  opened  and  the  points  readily 
cleaned  after  use.  The  ink  is  put  between  the  blades 
with  a  common  pen,  or  sometimes  by  a  small  hair 
brush.  In  using  the  drawing  pen  it  should  be  slightly 
inclined  in  the  direction  of  the  line  to  be  drawn,  and 
should  be  kept  uniformly  close  to  the  ruler  or  straiglit- 


Fig.  IS?.— Pencil  Sharpened  to  a  Round  Point, 


Fig.  123.— Ruling  Pen. 

edge  during  the  whole  operation  of  drawing  a  line, 
but  not  so  close  as  to  prevent  both  points  from  touch- 
ing the  paper  equally. 

Keeping  the  blades  of  the  pen  clean  is  essential 
to  good  work.  If  the  draftsman  is  careless  in  this  par- 
ticular, the  ink  will  soon  corrode  the  points  to  such  an 
extent  that  it  will  be  impossible  to  draw  fine  lines. 

Pens  will  gradually  wear  away,  and  in  course  of 
time  they  require  dressing.  To  dress  up  the  tips  of 
the  blades  of  a  pen,  since  they  are  generally  worn 
unequally  by  customary  usage,  is  a  matter  of  some 
nicety.  A  small  oil  stone  is  most  convenient  for  use 
in  the  operation.  The  points  should  be  screwed  into 
contact  in  the  first  place,  and  passed  along  the  stone, 
turning  upon  the  point  in  a  directly  perpendicular  plane 
until  they  acquire  an  identical  profile.  Xext  they  arc 
to  be  unscrewed  and  examined  to  ascertain  the  parts 
of  unequal  thickness  around  the  nib.  The  blades  are, 
then  to  be  laid  separately  upon  their  backs  upon  the 
stone,  and  rubbed  down  at  the  points  until  they  are 
brought  up  to  an  edge  of  uniform  fineness.  It  is  well 
to  screw  them  together  again  and  pass  them  over  the 


Tun/.*  mill  Materials. 


27 


stone  once  or  twice  more  to  bring  up  anv  fault  ami  t<> 
retouch  them  also  at  tin1  outer  and  inner  side  of  each 
blade  to  remove  barbs  or  framing,  and  linally  to  draw 
them  across  the  palm  of  the  hand. 

India  Ink. — .For  tracing*,  and  fur  some  kinds  of 
drawings,  which  the  ])att(irn  cutter  is  obliged  to  make 
occasionally.  India  ink  is  niueh  better  than  the  pencil, 
which  is  used  for  the  greater  part  of  his  work,  ('arc 
is  to  lie  exercised  in  the  selection  of  ink,  as  poor  grades 
arc  sold  as  well  as -good  ones.  Some  little  skill  is 
required  in  dissolving  or  mixing  it  for  use. 

India  ink  is  sold  in  cakes  or  sticks,  of  a  variety 
of  shapes.  It  is  prepared  for  use  by  rubbing  the1  end 
of  the  stick  upon  the  surface  of  aground  glass,  or  of  a 
porcelain  slab  or  dish,  in  a  very  small  quantity  of 
water,  until  the  mixture  is  siitlieiently  thick  to  produce 
a  black  line  as  it  Hows  from  the  point  of  the  ruling 
pen.  Tin1  qualitv  of  ink  may  generally  be  determined 
bv  the  price.  The  common  si/e  sticks  are  about  3 
inches  long.  Inferior  grades  can  be  bought  as  low  as 
40  cents  per  stick,  while  a  good  quality  is  worth  $(1.50 
to  *_'  per  stick,  and  the  very  best  is  still  higher.  How- 
ever, except  iu  the  hands  of  a  responsible  and  expe- 
rienced dealer,  this  method  of  judging  is  hardly 
satisfactorv.  To  a  certain  extent  ink  may  be  judged 
by  the  brands  upon  it,  although  in  the  ease  of  the 
higher  qualities  the  brands  frequently  change,  so  that 
this  test  may  not  be  infallible.  The  quality  of  India 
ink  is  quite  apparent  the  moment  it  is  used.  The  best 
is  entire!  v  free  from  grit  and  sediment,  is  not  musky, 
and  has  a  soft  feel  when  wetted  and  smoothed.  The 
color  of  the  lines  may  also  be  used  as  a  test  of  quality. 
\Vith  a  poor  ink  it  is  impossible  to  make  a  black  line. 
It  will  be  brown  or  irregular  in  color  and  will  present 
an  irregular  edge,  as  though  broken  or  ragged,  while 
an  ink  of  satisfactory  quality  will  produce  a  clean  line, 
whether  drawn  very  fine  or  quite  coarse. 

Various  shaped  cups,  slabs  and  dishes  are  in  use 
for  mixing  and  containing  India  ink.  In  many  re- 
spects thev  arc  like  those  used  for  mixing  and  holding 
water  colors.  Indeed,  in  many  cases  the  same  articles 
are  employed.  The  engraving  (Fig.  1'24)  shows  what 
is  termed  an  India  ink  slab,  with  three  holes  and  one 
slant.  This  article  is  in  common  use  among  draftsmen 
and  serves  a  satisfactory  purpose.  In  order  to  retard 
evaporation,  a  kind  of  saucers,  in  sets,  is  frequently 
used,  so  constructed  that  one  piece  will  form  a  cover  to 
the  other,  and  which  are  known  in  the  trade  as  cabinet 
sets  or  cabinet  saucers.  They  are  from  2£  to  3£ 
inches  in  diameter  and  come  in  sets  of  six.  In  the 


absence  of  ware  especially  designed  for  the  purpose, 
India  ink  can  be  satisfactorily  mixed  in  ;ind  used  from 
an  ordinary  saucer  or  plate  of  small  size.  The  articles 
made  especially  for  it,  however,  are  convenient,  and 


Front  with  Cover  On. 


Top  with  Cover  Off. 
Fig.  U4.—  India  Ink  Slab. 

in  facilitating  the  care  and  economical  use  of  the  ink 
are  well  worth  the  small  price  they  cost. 

Several  makes  of  liquid  drawing  ink  are  also  to 
be  had,  which  possess  the  advantage  of  being  always 
ready  for  use,  thus  doing  away  with  the  rubbing 
process.  The  ink  costs  about  25  cents  a  bottle, 
keeps  well,  and  will  answer  almost  every  purpose 
quite  as  well  as  the  stick  ink. 

Thumb  Tacks  or  Drawing  Pins,  both  names  being 
in  common  use,  are  made  of  a  variety  of  sizes,  ranging 
from  those  with  heads  one-quarter  of  an  inch  in  diam- 


Fig.  125.—  Thumb  Tacks,  or  Drawing  Pins. 

eter  up  to  eleven-sixteenths  of  an  inch  in  diameter. 
They  are  likewise  to  be  had  of  various  grades  and 
qualities.  The  best  for  general  use  are  those  of  Ger- 
man silver,  about  three-eighths  to  five-eighths  of  an 
inch  in  diameter,  and  with  steel  points  screwed  in  and 
riveted  Those  which  have  the  points  riveted  only 
are  of  the  second  quality.  The  heads  should  be  flat,  to 
allow  the  "T-square  to  pass  over  them  readily.  In  the 


28 


The  New  Metnl    \\'<>rke,-    Pattern  llonlc. 


annexed  cut.  Fig.  125,  are  shown  an  assortment  of 
kinds  and  sixes.  Those  which  are  beveled  upon  their 
upper  edges  are  preferable  to  those  which  are  beveled 
underneath. 

A  Box  of  Instruments.— Fig-  126  shows  a  box  of  in- 
struments of  medium  grade,  as  made  up  and  sold  by 
the  trade  general Iv.  While  it  contains  some  pieces 
that  the  pattern  cutter  has  no  use  for,  it  also  contains 
the  principal  tools  he  requires,  all  put  together  in  com- 
pact shape,  and  in  a  convenient  manner  for  keeping 
the  instruments  clean  and  in  good  order.  The  tray  of 
the  box  lifts  out,  there  being  a  space  underneath  it  in 
which  may  be  placed  odd  tools,  pencils,  etc.  Tools 
may  be  selected,  as  required,  of  most  of  the  large 
dealers  in  drawing  instruments.  It  will  be  found  ad- 
vantageous to  the  pattern  cutter  to  buy  his  instruments 
singly  as  he  requires,  them,  as  by  so  doing  he  will  get 
only  what  he  requires  for  use,  and  will  probably  secure 


Fig.  126. — A  Box  of  Instruments. 

a  better  quality  in  the  tools.  After  he  has  made  his 
selection,  a  box  properly  fitted  and  lined  should  be 
provided  for  them  and  can  be  obtained  at  a  small  cost, 
or  made  if  desirable. 

India  Rubber.— A  good  rubber  with  which  to  erase 
erroneous  lines  is  indispensable  in  the  pattern  cutter's 
outfit.  The  several  pencil  manufacturers  have  put 
their  brands  upon  rubber  as  well  as  upon  pencils,  and 
satisfactory  quality  can  be  had  from  any  of  them.  The 
shape  is  somewhat  a  matter  of  choice,  flat  cakes  being 
the  most  used.  A  very  soft  rubber  is  not  so  well 
adapted  to  erasing  on  detail  paper  as  the  harder  varie- 
ties, but  is  to  be  preferred  for  use  in  fine  drawings  on 
good  quality  paper. 

Paper. — The  principal  paper  that  the  pattern  cutter 
has  anything  to  do  with  is  known  as  brown  detail  paper, 
or  manila  detail  paper.  It  can  be  bought  of  almost 


any  width,  from  30  inches  up  to  54  inches,  in  rolls  of 
50  to  100  pounds  each.  It  is  ordinarily  sold  in  (lie 
roll  by  the  pound,  but  can  be  bought  at  retail  by  the 
yard,  although  at  a  higher  figure.  There  are  dilVerent 
thicknesses  of  the  same  qualitv.  Sonic  dealers  indi- 
cate them  by  arbitrary  marks,  as  XX,  XXX.  XXXX; 
others  by  numbers  1,  2,  3;  and  still  others  as  thin, 
medium  and  thick.  The  most  desirable  paper  for  tin- 
pattern  cutter's  use  is  one  which  combines  several  good 
qualities.  It  should  be  just  as  thin  as  is  consistent 
with  strength.  A  thick  paper,  like  a  still'  card,  breaks 
when  folded  or  bent  short,  and  is,  therefore,  objection- 
able. The  ] taper  should  be  very  strong  and  tough,  as 
the  requirements  in  use  arc  quite  severe.  The  surface 
should  be  very  even  and  smooth,  yet  not  so  glossv  as 
to  be  unsuitcd  to  the  use  of  hard  pencils.  It  should 
be  hard  rather  than  soft  and  should  be  of  such  a 
texture  as  to  withstand  repeated  erasures  in  the  same 
spot  without  damage  to  the  surface. 

White  drawing  paper,  which  the  pattern  cutter 
has  occasionally  to  use  in  connection  with  his  work. 
can  be  had  of  almost  every  conceivable  grade  and  in  a 
variety  of  sixes.  The  very  best  qualitv,  and  the  kinds 
suited  for  the  finest  drawing*,  come  in  sheets  exclu- 
sively, although  the  cheaper  kinds  are  also  made  in 
the  shape  of  sheets  as  well  as  in  rolls.  White  drawing 
paper  in  rolls  can  be  bought  of  different  widths,  rang- 
ing from  36  to  54  inches,  and  from  a  verv  thin  grade 
up  to  a  very  heavy  article,  and  of  various  surfaces.  It 
is  sold  by  the  pound,  in  rolls  ranging  from  30  to  40 
pounds  each,  and  also  at  retail  by  the  yard.  A  kind 
known  as  eggshell  is  generally  preferred  bv  architec- 
tural draftsmen. 

Drawing  paper  in  sheets  is  sold  by  the  quire,  and 
at  retail  by  the  single  sheet.  The  sixes  are  generally 
indicated  by  names  which  have  been  applied  to  them. 
The  following  are  some  of  the  terms  in  common  use. 
with  the  dimensions  which  they  represent  placed  op- 
posite : 


Cap. 


13  x  17  I  Elephant 23  x  28 

Demy 15  x  20   Atlas 21!  x  34 


17  x  22 


Columbier 23  x  35 


Medium 

Eoyal 19  x  24  |  DoubleElephant.   27  x  40 

Super  Royal 1!>  x  27    Antiquarian 31  x  53 

Imperial 22  x  30iKmpcror 48  x  <is 

Still  another  set  of  terms  is  used  in  designating  French 
drawing  papers.  Different  qualities  of  paper,  both  as 
regards  thickness,  texture  and  surface,  can  be  had  of 
any  of  the  sizes  above  named. 


Drawing   Tunis  «//</  .Mni'-ri<tk. 


29 


Tracing  Paper  and  Tracing  Cloth.— Tin-  pattern 
cutter  has  frequent  use  for  tracing  paper,  and  ti  good 
article,  which  combines  strength,  transparency  and 
suitable  surface,  is  very  desirable.  Tracing  paper  is 
sold  both  in  sheets,  in  size  to  correspond  to  the  draw-. 
ing  papers  above  described,  and  in  rolls,  to  correspond 
in  width  to  the  roll  drawing  paper.  It  is  usually 
priced  by  the  quire  and  by  the  roll,  although  single 
sheets  or  single  yards  are  to  be  obtained  at  retail. 
The  rolls,  according  to  the  kinds,  contain  from  20  to 
30  yards.  There  are  various  manufacturers  of  this 
article,  but  it  is  usually  sold  upon  its  merits,  rather 
than  by  any  brand  or  trade-mark.  Tracing  cloth,  or 
tracing  linen,  is  used  in  place  of  tracing  paper  where 
great  strength  aiid  durability  are  required.  This 


article  comes  exclusively  in  rolls,  ranging  in  width 
from  18  to  ±'2  inches.  There  are  generally  24-  yards 
to  the  roll,  and  prices  are  made  according  to  the. 
width,  or,  in  other  words,  according  to  the  superficial 
contents  of  the  roll.  Two  grades  are  usually  sold,  the 
lirst  being  glazed  on  both  sides  and  suitable  onlv  for 
ink  work,  and  the  second  on  but  one  side,  the  other 
being  left  dull,  rendering  it  suitable  for  pencil  marks. 
Upon  general  principles,  pencil  marks  are  not  satis- 
factory upon  cloth,  even  upon  the  quality  specially 
prepared  with  reference  to  them.  It  is  but  a  very 
little  more  labor  or  expense  to  use  ink,  and  a  much 
more  presentable  and  usable  drawing  is  made.  Tracing 
paper  may  be  used  satisfactorily  with  either  pencil  or 
pen. 


CHAPTER  III. 


In  the  production  of  all  great  constructive  works 
the  drawing  plays  a  most  important  part.  If  a  piece 
of  machinery,  a  ship,  an  aqueduct  or  a  temple  is  to  be 
built,  verbal  descriptions  would  be  insufficient  direc- 
tions to  the  workmen  who  are  to  perform  the  actual 
labor;  drawings  become  a  necessity,  because  a  draw- 
ing tells  exactly  what  is  meant,  where  words  would 
utterly  fail.  Therefore,  to  everybody  connected  with 
the  constructive  trades,  to  artisans  in  whatever  field, 
the  ability  to  read,  if  not  to  make,  a  drawing  becomes 
a  necessity;  and  to  those  in  positions  of  authority  the 
ability  to  make  a  drawing  is  the  power  to  convey  their 
ideas  to  others.  That  branch  of  drawing  with  which 
the  pattern  cutter  has  to  deal  is  of  a  purely  geomet- 
rical nature  and  is  properly  termed  orthographic  projec- 
tion. The  term  orthographic  (signifying  right  line)  is 
well  applied  because  it  exactly  describes  the  nature  of 
the  work,  as  will  be  seen  further  on. 

The  geometrical  drawings  made  use  of  in  repre- 
senting any  constructive  work,  whether  to  a  large  or  a 
small  scale,  are  of  three  kinds — viz.  :  ELEVATIONS, 
SECTIONS  and  PLANS.  The  term  diagram  is  sometimes 
used  in  connection  with  this  class  of  drawing,  but  is 
not  of  a  specific  nature.  It  means  a  drawing  of  the 
simplest  possible  character,  usually  made  to  demon- 
strate a  principle,  and  may  partake  of  the  properties  of 
either  of  the  above  named  drawings. 

An  elevation,  if  the  word  were  judged  by  its  com- 
mon meaning,  would  be  understood  to  show  the  hight 
of  anything.  It  does  this  and  more.  It  gives  all  the 
vertical  and  horizontal  measurements  which  appear  in 
the  front,  side  or  end  which  it  represents.  An  elevation 
supposes  the  observer  to  be  opposite  to  and  on  a  level 
with  all  points  at  the  same  time,  and  is  therefore  an 
impossible  view,  according  to  the  rules  of  pictorial  art. 
Being  always  drawn  to  scale  (including  full  size),  it 
gives  exact  dimensions  of  hight  and  breadth  at  any 
part  of  the  view,  but  furnishes.no  view  of  horizontal 
surfaces  and  no  means  of  measuring  distances  to  HIP  I 


from  the  observer,  or  in  any  oblique  horizontal  direc- 
tion. An  elevation  may  be  called  front,  xide,  end  or 
rear,  according  to  the  relative  dimensions  of  the  object, 
one  of  whose  faces  it  represents.  Any  elevation  or 
vertical  section  gives  two  sets  of  dimensions — i.  e., 
hight  and  horizontal  distance,  which  lie  parallel  to  the 
face  which  it  shows. 

A  section,  as  the  word  imlie;ites.  is  a  view  of  a 
cut  or  a  view  of  what  remains  after  certain  portions 
have  been  cnt  away  for  the  purpose  of  showing  more 
clearly  the  interior  construction.  The  idea  of  a  verti- 
cal section  can  best  be  described  by  supposing  that  a 
wire  stretched  taut,  or  any  perfectly  straight  blade, 
was  passed  vertically  down  through  an  object  at  a  given 
distance  from  one  of  its  ends  or  sides,  indicated  by  a 
line  in  some  other  view  or  views,  and  the  portion  not 
wanted  was  removed.  The  view  made  of  the  section 
may  properly  include  only  the  parts  cnt,  or  if  made  to 
include  or  show  portions  that  would  naturally  appear 
by  the  removal  of  the  parts,  it  would  properly  be  called 
a  sectional  elevation.  Sections  may  also  be  taken  hori- 
zontally at  any  hight  above  the  base  or  ground  line, 
indicated  by  a  line  for  that  purpose  upon  one  or  more 
of  the  elevations. 

Horizontal  sections  are  properly  classed  with  plans. 
Vertical  sections  are  known  as  longitudinal  or  tranxri-w, 
according  as  they  are  taken  through  the  long  way  of, 
or  across,  an  object.  Elevations  or  sections  may  also 
be  constructed  upon  oblique  planes  when  necessary  to 
more  fully  show  construction. 

Sections  of  small  portions  or  members  drawn  to  a 
large  scale  or  full  size  are  called  profiles.  They  are  ap- 
plied to  continuous  forms,  as  moldings,  jambs,  etc., 
and  are  drawn  for  the  purpose  of  showing  the  peculiari- 
ties in  form  of  the  parts  which  they  represent. 

The  view  which  gives  all  the  horizontal  distances 
in  whatever  direction  is  called  the  plan.  The  name 
plan  applies  equally  well*  to  a  horizontal  section  or  to 
a  top  view.  In  the  plan,  as  in  sections  and  elevations, 


Drawing. 


31 


the  observer  is  supposed  to  be  opposite  to  (i.e.,  di- 
rectly above)  all  points  at  the  same  time.  In  idea  it 
is  the  same  as  a  map,  the  difference  between  the  two 
terms  being  in  the  amount  included  in  the  view. 

In  Fig.  1-<S  is  given  an  illustration  of  the  various 
geometrical  views  of  an  object,  placed  in  their  propel' 
relation  one  to  another,  showing  the  lines  of  projec- 
tion and  the  lines  upon  which  the  different  sections 
are  taken.  A  house  placed  upon  a  base  has  been  se- 
lected as  the  most  suitable  object  for  purposes  of 
explanation  in  the  present  case.  It  has  been  shown 
'in  diagrammatic  form — that  is,  denuded  of  all  cornices, 
trimmings  or  projecting  parts — so  as  to  demonstrate 
the  principles  of  projection  in  the  clearest  manner 
possible. 

It  rests  with  the  designer  to  determine  which  of 
the  views  shall  be  drawn  first,  all  depending  upon  the 
given  facts  or  specifications  in  his  possession.  If  a 
house  is  to  be  designed,  it  is  most  likely  that  the  plan 
would  be  drawn  first,  as  arrangement  of  rooms  and 
amount  of  ground  to  be  covered  would  be  of  the  first 
importance.  If  a  molding  be  the  subject  of  the  de- 
sign, the  profile  would  be  the  view  in  which  to  first 
adjust  the  proportion  of  its  parts.  The  method  of 
deriving  the  elevation  from  the  section  or  obtaining 
any  one  view  from  one  or  more  other  views  is  termed 
orthographic  projection,  because  by  it  a  system  of 
parallel  lines  is  made  use  of  for  the  purpose  of  obtain- 
ing the  same  hight  (or  width,  as  the  case  may  be)  in 
corresponding  parts  in  the  different  views, 

In  this  connection  it  is  to  be  understood  that  each 
angle  or  limit  of  outline  in  a  sectional  view  is  the 
source  of  a  right  line  in  the  elevation.  In  Fig.  127  is 
shown,  at  X,  a  sectional  view  or  profile  of  a  molding, 
which  should  be  so  drawn  that  all  the  faces  or  sur- 
faces supposed  to  be  vertical  shall  lie  vertically  on  the 
paper;  that  is,  parallel  to  the  sides  of  the  drawing 
board.  To  project  an  elevation,  Y,  from  this  section, 
place  the  T-square  so  that  the  blade  lies  horizontal— 
that  is,  crossing  the  board  from  side  to  side — and  bring" 
it  to  the  various  angles  A,  B,  C,  etc.,  of  the  profile, 
drawing  a  line  from  each.  The  point  E,  though  not 
an  angle,  is  the  lowest  visible  point  or  limit  of  that 
member  of  the  mold  when  seen  from  the  front,  and  is, 
therefore,  entitled  to  representation  in  the  elevation 
by  a  line.  Tn  like  manner  the  point  T>,  being  the 
upper  limit  of  a  curve,  is  entitled  to  representation. 
but  being  so  situated  as  to  be  invisible  when  viewed 
from  a  point  in  front  of  the  mold,  the  line  is  properly 
made  dotted,  The  lines  of  projection  from  the  section 


to  the  elevation  are  also  shown  dotted  in  the  engrav- 
ing. A  vertical  line  terminates  the  elevation  of  the 
mold  at  the  right  or  end  nearest  the  section,  while  the 
absence  of  such  a  line  at  its  left  end  indicates  that  it 
extends  indefinitely  in  that  direction.  It  would  also 
be  proper,  upon  that  supposition,  to  finish  the  eleva- 
tion at  the  left  with  a  broken  line. 

Referring  now  to  Fig.  I'-'*,  it  is  most  likely  that 
the  front  elevation  would  be  next  drawn  after  the  plan. 
For  this  purpose  the  plan  should  be  so  placed  upon 
the  board  that  the  part  representing  the  front  should 
be  turned  toward  the  bottom  of  the  board,  in  which 
position  it  appears  to  be  turned  toward  the  observer. 
Place  the  "["-square  so  that  the  blade  lies  vertically 
upon  the  board — that  is,  crossing  it  from  front  to 
back — and  bringing  it  to  the  different  angles  or 
points  of  the  front  side  of  the  plan,  draw  a  line  ver- 
tically from  each,  through  that  portion  of  space  upon 


Fig.  1S7.— Elevation  Projected  from  Section. 

the  paper  allotted  to  the  elevation,  all  as  shown  by  the 
dotted  lines.  Thus  each  point  of  the  elevation  comes 
directly  over  the  point  which  represents  it  in  the  plan, 
and  the  horizontal  distance  across  any  part  of  the  new 
elevation  thus  becomes  exactly  the  same  as  that  of  the 
plan.  The  question  of  bights  is  here  a  matter  of  de- 
sign and  is  governed  by  specifications  supplemented 
by  the  designer's  judgment.  With  the  plan  and  the 
front  elevation  complete  the  drawing  of  any  other  ele- 
vations or  sections  is  entirely  a  matter  of  projection, 
except  as  new  features  might  occur  in  those  views 
which  would  not  appear  in  either  of  the  views  already 
drawn. 

If  an  elevation  of  the  right  side  is  about  to  be  con- 
structed, lines  would  be  projected  horizontally  to  the 
right  from  every  point  in  the  front  elevation  of  the  ob- 
ject which  would  be  visible  when  seen  from  the  right 
side,  thus  locating  all  the  hights  in  the  new  view.  As 
the  horizontal  distances  in  this  view  must  agree  with 
distances  from  front  to  back  on  the  plan,  they  may  best 
lie  obtained  by  turning  the  plan  (or  so  much  of  itas  nec- 
essarv  to  this  view)  one-quarter  around  to  the  right,  so 


32 


The  Sew  Metal    \\~ui-ker   I'atttni    il<><>k. 


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Linear   Drawing. 


33 


tluit  tlic  side  of  \vliidi  the  new  elevation  is  1<>  lie  drawn 
will  he  toward  the  bottom  'or  near  side,  o[  the  lioanl. 
as  shown  at  (i  :  after  which  lines  inav  lie  projected  with 
the  T-square  from  the  points  of  the  plan  into  the  eleva- 
tion, intersecting  with  corresponding  lines,  as  shown. 
The  same  result  ma\  he  accomplished  hv  projecting 
the  lines  to  the  right  from  the  side  of  the  plan,  as 
shown  in  the  top  view,  until  thev  reach  anv  line  paral- 
lel to  the  side,  as  II  I.  PYoin  this  line  tliev  may  be 
carried  around  a  quarter  circle  from  any  convenient 
center,  as  N,  arriving  at  a  horizontal  line,  N  M,  and 
thence  dropped  downward,  intersecting  as  before. 

It  will  thus  he  seen  that  the  elevation  of  the  right 
hand  side  of  anv  object  comes  naturally  at  the  right  of 
the  front  elevation,  and  the  left  side  elevation,  at  its 
left.  This  idea  is  best  illustrated  hv  supposing  that 
the  ohject  in  question  he  placed  in  a  glass  box  of  the 
dimensions  of  the  base  II  I  J  K  of  the  top  view,  and 
that  thi'  elevation  of  cadi  side  of  the  object  be  pro- 


LONGITUDINAL  SECTION  ON  C-D  TRANSVERSE  SECTION  ON  A-B 

Fig.  129.— Vertical  Sections  Derived  from  Fig.  128. 


jected  upon  the  adjacent  parallel  side  of  the  box  at 
right  angles  to  the  same,  and  that  afterward  all  the 
sides  (supposing  them  to  be  hinged  at  the  corners)  be 
opened  out  into  one  plane,  as  shown  by  K  L,  II  O  and 
O  1'  (the  top  face  of  the  box  being  opened  upward), 
thus  displaying  all  the  views  in  one  plane  as  repre- 
sented by  Fig.  12S. 

This  idea  should  not  be  carried  so  far  as  to  open 
the  bottom  face  of  the  box  downward,  because  this 
would  produce  a  plan  as  seen  from  below,  which  is 
never  done  except  in  the  case  of  a  design  of  a  ceiling 
or  sollit.  when  it  should  be  spoken  of  as  an  inverted 
plan. 

In  Fig.  129  the  transverse  section  is  shown  at  the 
right  of  the  longitudinal  section,  because  the  view  in 
it  is  from  the  right,  or  in  the  direction  of  the  arrow  in 
the  longitudinal  section,  showing  what  would  he  seen 
if  the  house  were  cut  in  two  on  the  line  A  B  of  the 
plan  and  the  right  hand  portion  removed.  The  longi- 


tudinal section  is  for  the  same  reason  placed  at  the  left, 
of  the  transverse  section — that  is,  it  is  a  view  from  the 
left  of  the  house  when  placed  in  the'  position  shown  hv 
the  transverse  section.  From  the  foregoing  it,  is  to  he 
understood,  therefore,  that  when  a  view  appears  to  the 
ri^ht  of  another  it  is  supposed  to  show  what  would  be 
seen  when  the  object  is  viewed  from  the  right  hand 
end  or  side  of  what  is  shown  in  the  other,  the  other 
(or  front)  view  being  at  the  same  time  a  view  of  the 
left  side  of  what  is  shown  by  the  right  side  elevation. 

In  this  class  of  drawings  various  kinds  of  lines 
are  used,  cadi  of  which  possesses  a  certain  significance. 
The  general  outlines  of  the  different  views  should 
be  linn  and  strong  enough  to  he  distinctly  visible,  with- 
out being  so  broad  as  to  leave  any  doubt  as  to  the  ex- 
ad  dimensions  of  the  part  shown  when  the  rule  is  ap- 
plied for  purposes  of  measurement. 

It  is  not  always  necessary  that  all  the  lines  of 
projection  should  be  shown.  When  shown  they  may 
appear  as  the  finest  possible  continuous  lines, 
or  as  dotted  lines  such  as  are  shown  in  Figs. 
127,  128  and  129.  Lines  used  in  carrying 
points  from  a  profile  to  a  miter  line,  or  from 
one  line  to  another  for  any  purpose,  are  rea  1 1  v 
lines  of  projection,  and  for  the  pattern  drafts- 
man's purposes  it  may  be  said  that  the  liner 
they  are  drawn  the  greater  the  accuracy  ob- 
tained (see  Chapter  II  under  the  head  of  Lead 
Pencils). 

Dotted  lines  are  also  used  to  represent 
portions  which  are  out  of  sight — that  is,  back 
of  or  underneath  the  other  parts  which  constitute  the 
view  under  consideration,  but  which  it  is  necessary  to 
show,  as,  for  instance,  a  portion  of  the  chimnev  in 
longitudinal  section  Fig.  129  and,  points  1)  and  F  in 
the  profile  of  the  mold  in  Fig.  127. 

Dotted  lines  are  also  used  to  show  a  change  of 
position  or  an  alternate  position  of  some  part,  as,  for 
example,  the  lines  L  K  and  .J  L  show  that  the  side  .1  K 
of  the  top  view  has  been  swung  around  on  the  point  K 
until  it  occupies  the  position  shown  by  L  K,  its  ex- 
tremity J  traversing  the  line  J  L.  When  it  is  neces- 
sary to  use  two  kinds  of  dotted  lines,  those  used  for 
one  purpose  may  be  made  in  line  or  short  dots,  while 
the  others  may  be  made  a  series  of  short  dashes. 

Lilies  showing  the  part  of  a  view  through  which  a. 
section  is  taken  are  composed  of  a  series  of  dots  and 
dashes,  as  shown  by  A  B,  (.'  1>,  etc.,  in  Fig.  12S,  and 
when  further  distinction  is  required  may  be  made  by 
two  dots  alternating  with  a  short  or  long  dash. 


34 


The  Xcw   Metal    UW/yr    I \Mvrn   Book. 


When  it  is  desirable  to  omit  the  drawing  of  a  con- 
siderable portion  of  any  view  it  is  customary  to  termi- 
nate the  incomplete  side  of  such  view  by  an  irregular 
line,  as  shown  above  the  plan  G  in  Fig.  128. 

It  is  customary  in  all  sectional  views  for  the  parts 
which  are  represented  as  being  cut  to  be  ruled  or  lined 
with  lines  running  in  an  oblique  direction,  as  in  Fig. 
129.  When  the  section  comprises  several  different, 
pieces  lying  adjacent  to  one  another,  each  different 
part  should  be  lined  in  a  different  direction.  This  rul- 
ing is  understood  to  mean  solidity.  In  Fig.  129  the 
walls  and  base  in  the  different  sections  are  represented 
as  though  made  of  some  solid  material,  as  wood  or 
stone,  and  ruled  accordingly.  Where  it  is  necessary 
to  represent  different  kinds  of  material  in  the  same  sec- 
tion, different  systems  or  kinds  of  lines  may  be  used 
for  the  purpose.  Thus  solid  and  dotted  lines  may  be 
used  alternately,  as  in  the  base.  Coarse  and  fine  rul- 
ing, or  stippling,  may  also  be  employed,  according  to 
the  size  of  the  part,  or  very  small  parts  may  be  shown 
solid  black,  as  window  weights,  piping  or  hinges.  A 
heavy  line  is  the  only  way  that  a  thickness  of  metal 
can  properly  be  shown  in  a  section.  In  the  case  of  a 
sectional  view  of  a  cornice  or  molding  where  nothing 
but  the  sheet  iron  appears,  it  is  customary  to  make  use 
of  section  lines  close  to  the  metal  surface,  but  not  to 
extend  them  clear  across  the  space  which  should  be 
filled  if  the  moldings  were  of  stone  or  other  solid  mate- 
rial. By  this  means  a  section  may  be  distinguished 
from  what  might  otherwise  be  taken  for  an  elevation 
of  a  return. 

In  the  case  of  elaborate  drawings  prepared  by  an 
architect  color  is  frequently  resorted  to  as  a  means  of 
showing  the  different  materials  as  they  appear  in  the 
sectional  view,  yellow  or  differing  shades  of  brown 
being  used  for  various  kinds  of  wood,  while  blue  is 
generally  used  for  iron,  gray  for  stone,  red  for  brick, 
etc.  In  the  case  of  drawings  showing  many  different 
materials  it  is  usual  to  place  a  legend  in  one  corner  of 
the  drawing  showing  what  each  color  or  style  of  ruling 
indicates. 

It  is  always  advisable  to  keep  the  different  views, 
which  it  is  necessary  to  construct,  separate  and  dis- 
tinct from  one  another,  drawing  them  as  near  to- 
gether as  circumstances  will  permit,  but  never  allow- 
ing one  view  to  cover  any  part  of  the  space  upon  the 
paper  occupied  by  the  other  view  if  it  can  be  avoided. 
One  notable  exception  to  this  rule  is  to  be  observed. 
It  frequently  occurs  in  drawing  an  elevation  of  a  large 
surface,  as  a  pediment  or  side  of  a  bracket,  that  it  is 


necessary  to  indicate  that  some  part  of  it  is  recessed  or 
raised,  or  that  a  certain  edg'e  is  molded  or  chamfered, 
when  it  would  not  be  necessary  to  construct  an  entire 
sectional  view  for  this  purpose  alo;ie.  To  this  end  it 
is  customary  to  draw  through  such  mold,  chamfer  or 
recess  a  small  section,  in  which  case,  if  tin-  depression 
or  mold  runs  horizontally,  the  section  is  turned  i<>  the. 
right  or  left,  according  to  convenience,  <>r  if  it  runs 
obliquely,  it  is  turned  in  the  direction  the  mold  runs. 
In  such  a  section  the  line  which  represents  the.  plane 
surface  also  shows  the  direction  of  the  cut  across  the 
mold  or  line  upon  which  the  section  is  taken.  In  Fig. 
i:;n  is  shown  an  elevation  of  a  portion  of  a  pediment, 
in  which  a  small  section,  A  B  G,  is  introduced  to  show 
the  profiles  of  the  moldings.  The  line  B  C,  which 


Fig.  ISO,  —Elevation  with  Section  of  Parts. 


represents  the  profile  of  the  stile  around  the  panel, 
shows  the  line  upon  which,  or  the  direction  in  which, 
the  section  is  taken,  said  section  being  turned  upon 
this  line  obliquely  to  the  left.  It  is  necessary  to  rule 
or  line  this  section,  the  ruling  being  kept  close  to  and 
inside  the  outline  or  profile.  By  placing  the  ruling  in- 
side the  profile  no  doubt  can  exist  as  to  which  parts 
are  raised  and  which  are  depressed,  for  if  at  D  the 
ruling  were  upon  the  other  side  of  the  line  from  that 
shown  the  section  D  would  indicate  a  depressed  panel 
instead  of  a  raised  one. 

In  the  solution  of  the  class  of  problems  treated 
in  Chapter  VI,  Section  1  (Miter  Cutting),  confusion 
often  arises  in  the  mind  of  the  pattern  cutter  as  to  the 
proper  position  of  a  profile  or  of  a  miter  line,  which 


Linear    Drawing, 


35 


confusion  could  never  occur  if  all  the  necessary  views 
were  lirst  drawn  In  accordance  with  the  principles 
which  this  chapter  is  written  to  explain.  A  pro- 
file is  always  a  section,  and  a  miter  line  is  either  a  part 
of  an  elevation  projected  from  the  section  or  part  of 
another  section  bearing  certain  relations  of  hight  or 
breadth  to  the  first.  A  pattern  is  likewise  always  pro- 
jected— that  is,  carried  off  by  right  lines — from  an 
elevation  or  plan  the  same  as  an  elevation  is  projected 
from  a  section. 

Jt  should  also  be  remembered  in  this  connection 
that  the  operation  of  developing  a  pattern  is  not  com- 
pleted until  its  entire  outline  is  drawn.  The  line  form- 
ing its  termination  at  the  end  opposite  the  miter  cut, 


although  simply  u  straight  line,  is  properly  derived 
from  the  elevation  or  plan  used,  the  same  as  all  points 
and  other  lines  of  the  pattern. 

Much  trouble  is  experienced  through  lack  of 
knowledge  of  the  principles  of  Linear  Drawing,  which 
if  thoroughly  understood  could  never  result  in  such 
mistakes  as  producing  a  face  miter  where  a  return 
was  intended  or  using  the  piece  of  metal  from  the 
wrong  side  of  the  miter  cut. 

Too  much  emphasis  cannot  be  placed  upon  the  im- 
portance of  thoroughly  understanding  the  subject  treated 
in  this  chapter,  as  such  a  knowledge  comprehends 
within  itself  an  answer  to  the  many  questions  continually 
arising  in  the  course  of  the  pattern  draftsman's  labors. 


CHAPTER    IV. 


In  presenting  this  chapter  to  the  student  no  at- 
tempt has  been  made  to  give  a  complete  list  of  ^co- 
metrical  problems,  but  all  those  have  been  selected 
which  can  be  of  any  assistance  to  the  pattern  drafts- 
man, and  especial  attention  has  been  given  in  their 
solution  to  those  methods  most  adaptable  to  his  wants. 
They  are  arranged' as  far  as  possible  in  logical  order 
and  are  classified  under  various  sub-heads  in  such  a 
manner  that  the  reader  will  have  no  difficulty  in  finding 
what  he  wishes  by  simply  looking  through  the  pages, 
the  diagrams  given  with  each  being  sufficient  to  indi- 
cate the  nature  of  the  problem  and,  as  it  were,  form  a 
sort  of  index. 

1.  To  Draw  a  Straight  Line  Parallel  to  a  Given  Line 
and  at  a  Given  Distance  from  it,  Using  the  Compasses  and 

Straight-Edge.— In  Fig.  131,  let  C  D  be  the  given  line 
parallel  to  which  it  is  desired  to  draw  another  straight 
line.  Take  any  two  points,  as  A  and  B,  in  the  given 
line  as  centers,  and,  with  a  radius  equal  to  the  given 
distance,  describe  the  arcs  x  x  and  y  y.  Draw  a  line, 
touching  these  arcs,  as  shown  at  E  F.  Then  E  F  will 
be  parallel  to  C  D. 

2.  To  Draw  a  Line  Parallel  to  Another  by  the  Use  of 
Triangles  or  Set-Squares.— In  Fig.  132,  let  A  B  be  the 
line  parallel  to  which  it  is  desired  to  draw  another. 
Place  one  side  of  a  triangle  or  set-square,  F1,  against 
it,  as  indicated  by  the  dotted  lines.      While  holding 
F1  firmly  in  this  position,  bring  a  second  triangle,  or  any 
straightedge,  E,  against  out1  of  its  other  sides,  as  shown. 
Then,  holding  the  second  triangle  firmly  in  place,  slide 
the  first  away  from  the  given  line,  keeping  the  edges 
of  the  two  triangles  in  contact,  as  shown  in  the  figure. 
Against  the  same  edge  of  the   first  triangle  that  was 
placed  against  the  given   line  draw  a  second  line,  as 
shown  by  0  1).      Then  C  1)   will   be   parallel   to  A  B. 
In  drawing  parallel    lines    by  this   method   it  is  found 
advantageous  to   place  the    longest   edges  of    the   tri- 
angles against  cadi  other,  and  to  so  place   the   two    in- 
struments that  the  movement  of  one   triangle  against 


the  other  shall  be  in  a  direction  oblique  to  the  lines 
to  be  drawn,  as  greater  accuracy  is  attainable  in  this 
way. 

3.  To  Erect  a  Perpendicular  at  a  Given  Point  in  a 
Straight  Line  by  Means  of  the  Compasses  and  Straight- 

Edge. — In  Fig.  133,  let  A  B  represent  the  given 
straight  line,  at  the  point  C  in  which  it  is  required  to 
erect  a  perpendicular.  From  C  as  a  center  with  any 
convenient  radius  strike  small  arcs  cutting  A  B,  as 
shown  \)y  D  and  B.  With  D  and  B  as  centers,  and 
with  any  radius  longer  than  the  distance  from  each  of 
these  points  to  C,  strike  arcs,  as  shown  by  x.rand  //  //. 
From  the  point  at  which  these  ares  intersect,  E,  draw  a 
line  to  the  point  C,  as  shown.  Then  E  C  will  be  per- 
pendicular to  A  B. 

4.  To  Erect  a  Perpendicular  at  or  near  the  End  of 
a  Given  Straight  Line  by  Means  of  the  Compasses  and 
Straight-Edge.— First  Method.— In  Fig.  134,  let  A  B  be 
the  given  straight  line,  to  which,  at  the  point  P,  situ- 
ated near  the  end,  it  is  required  to  erect  a  perpendic- 
ular.     Take  any  point  (C)  outside  of  the  line  A  B. 
With  C  as  center,   and   with   a    radius   equal     to    thr 
distance  from  C  to  P,  strike  the  arc,  as  shown,  cutting  the 
given  line  A  B  in  the  point  P,  continuing  it  till  it  also 
cuts  in  another  point,  as  at   E.      From  E,  through    the 
center,  C.  draw  the  line  E  F,  cutting  the  arc,  as  shown 
at  F.      Then  from  the  point  F,  thus  determined,  draw 
a  line  to  P,  as  shown.      The   line  F  P  is  perpendic  nlar 
to  A  B. 

5.  To  Erect  a  Perpendicular  at  or  near  the  End  of 
a  Given  Straight  Line  by  Means  of  the  Compasses  and 
Straight-Edge.— Second  Method.— In  Fig.  135,  let  B  A 
be  the  given  straight  line,  to  which,  at  the  point  P,  it  is 
required  to  erect  a  perpendicular.      From  the  point  1', 
witli   a   radius   equal    to   three   parts,  by  any  scale,  de- 
scribe an   arc,  as   indicated    by   ./•  ,r.       From    the   same 
point,  with   a   radius   equal  to  four  parts,  cut    the   line 
B  A  in  the  point  C.      From  the  point  0,  witli  a  radius 
equal  to  live  parts,  intersect  the  arc  first  drawn  by  the 


Geometrical   Problems. 


87 


arc  y  ?/.  From  the  point  of  intersection  T)  draw  the  line 
1)  P.  Then  I)  P  will  l.e  perpendicular  to  H  A. 

6.  To  Draw  a  Line  Perpendicular  to  Another  Line 
by  the  Use  of  Triangles  or  Set  Squares. — In  Fig.    1  .">»;. 

let  ('  I)  be  the  given  line,  perpendicular  to  which  it  is 
required  to  draw  another  line.  Place  one  side  of  a 
triangle.  R,  against  the  given  line,  as  shown.  Rring 
another  triangle,  A,  or  any  straight  edge,  against  tin- 
long  side  or  hypothenuse  of  the  triangle  R,  as  shown. 
Then  move  the  triangle  R  along  the  straight  edge  or 
triangle  A,  as  indicated  hv  the  dotted  lines,  until  the 
opposite  side  of  R  crosses  the  line  (!  1)  at  the  required 


sides  of  the  given  line  A  B.  A  line  drawn  through 
these  points  of  intersection,  as  shown  by  G  H,  will 
liiseet  the  line  A  15.  or,  in  other  words,  divide  it  into 
two  equal  parts. 

8.  To  Divide  a  Straight  Line  into  Two  Equal  Parts 
by  the  Use  of  a  Pair  of  Dividers. — In  Fig.  13S,  it  is  re- 
quired to  divide  the  line  A  R  into  two  equal  parts,  or 
to  lind  its  middle  point.  Open  the  dividers  to  as  near 
half  of  the  given  line  as  possible  by  the  eye.  Place 
one  point,  of  the  dividers  on  one  end  of  the  line,  as  at 
A.  Rring  the  other  point  of  the  dividers  to  the  line, 
as  at  C,  and  turn  on  this  point,  carrying  the  first 


/•',•;/.  ;?/  —To  Draw  a  Straight  Line  Par- 
allel to  a  Given  Straight  Line,  and  at  a 
d'ifeti  Distance  from  it,  Using  the,  Com- 
j>ttfsr,es  and  a  Straight-Edge. 


Fig.  1.12.— To  Draw  a  Line  Parallel  to 
Another  by  the  Use  of  Triangles  or 
Set  Squares. 


Fig.  1S3  —To  Erect  a  Perpendicular  at 
a  Given  Point  in  a  Straight  Line, 
Using  the.  Compasses  and  Straight- 
Edgt. 


\ 


<  9 


•'iy.  1;'4  —To  Erect  a  perpendicu- 
lar at  or  near  the  End  of  a 
Given  Straight  Line,  Using  the 
Compassi  x  anil  Straight-Edge. 
First  Method. 


D/ 
^~ 


x 


X 


P  1  Parts  C 

Fig.  ISS.  —  To  Erect  a  Perpendicular 
at  or  near  the  End  of  a  Given 
Straight  Line.  Second  Method. 


Fig.  1X6.— To  Draw  a  Line  Perpen- 
dicular to  Another  by  the  Use  of 
Triangles. 


point.  When  against  it,  draw  the  line  E  F,  as  shown. 
Then  K  F  is  perpendicular  to  C  D.  It  is  evident  that 
this  rule  is  adapted  to  drawing  perpendiculars  at  anv 
point  in  the  given  line,  whether  central  or  located  near 
the  end.  Its  use  will  be  found  (-specially  convenient 
for  erecting  perpendiculars  to  lines  which  run  oblique 
to  the  sides  of  the  drawing  board. 

7.  To  Divide  a  Given  Straight  Line  into  Two  Equal 
Parts,  with  the  Compasses,  by  Means  of  Arcs.— In  Fig.  137, 
let  it  be  required  to  divide  the  straight  line  A  R  into  two 
equal  parts.  From  the  extremes  A  and  R  as  centers,  and 
with  any  radius  greater  than  one-half  of  A  R,  describe 
the.  arcs  rf/and  a  e,  intersecting  each  other  on  opposite 


around  to  D.  Should  the  point  D  coincide  with  the 
other  end  of  the  line,  the  division  will  be  correct. 
Rut  should  the  point  I)  fall  within  (or  without)  the 
end  of  the  line,  divide  this  deficit  (or  excess)  into  two 
equal  parts,  as  nearly  as  is  possible  by  the  eye,  and 
extend  (or  contract)  the  opening  of  the  dividers  to  this 
point  and  apply  them  again  as  at  first.  Thus,  finding 
that  the  point  D  still  falls  within  the  end  of  the  line,  the 
first  division  is  evidently  too  short.  Therefore,  divide 
the  deficit  I)  R  by  the  eye,  as  shown  by  E,  and  in- 
crease the  space  of  the  dividers  to  the  amount  of  one 
of  D  E.  Then,  commencing  again  at  A,  step  off  as 
before,  and  finding  that  upon  turning  the  dividers 


38 


'Hie    New 


Worker    PnW'rn    Honk-. 


upon  the  point  F  the  other  point  coincides  with  the 
end  of  the  line  B,  F  is  found  to  be  the  middle  point  in 
the  line.  In  some  cases  it  may  be  necessary  to  repeat 
this  operation  several  times  before  the  exact  center  is 
obtained. 

9.  To  Divide  a  Straight  Line  into  Two  Equal  Parts 
by  the  Use  of  a  Triangle  or  Set  Square.— In  Fig.  139, 
let  A  B  be  a  given  straight  line.  Place  a  J-square 


/ 

\ 

/ 

\ 

\ 

1 
1 
1 

\ 
\ 
\ 
\ 
I 

t 

1 

1 

1 

I                   R 

i 

i 

\ 

i 

\ 
\ 

\ 

i 
f 

\ 

t 

\ 

/ 

\ 

\ 

\ 

e'H-f 

Pig.  1S7.—TO  Divide   a  Straight  Line  into  Two  Equal  Parts 
by  Means  of  Arcs. 

or  some  straight  edge  parallel  to  A  B.  Then  bring 
one  of  the  right-angled  sides  of  a  set  square  against  it, 
and  slide  it  along  until  its  long  side,  or  hypothenuse, 
meets  one  end  of  the  line,  as  A.  Draw  a  line  along 
the  long  side  of  the  triangle  indefinitely.  Keverse  the 
position  of  the  set  square,  as  shown  by  the  dotted  lines, 
bringing  its  long  side  against  the  end,  B,  of  the  given 


Fig.  lS8.—To  Bisect  a  Straight  Line    by   the   Use  of  the 
Dividers. 

t 

straight  line,  and  in  like  manner  draw  a  line  along  its 
long  side  cutting  the  first  line.  Next  slide  the  set 
square  along  until  its  vertical  side  meets  the  intersec- 
tion of  the  two  lines,  as  shown  at  C,  from  which  point 
drop  a  perpendicular  to  the  line  A  B,  cutting  it  at  D. 
Then  D  will  be  equidistant  from  the  two  extremities 
A  and  B. 

10.  To  Divide  a  Given  Straight  Line  into  Any  Num- 
ber of  Equal  Parts.— In  Fig.  140,  let  A  B  be  a  given 
straight  line  to  be  divided  into  equal  parts,  in  this  case 


eight.  From  one  extremity  in  this  line,  as  at  A,  draw 
a  line,  as  cither  A  C  or  A  D.  oblique  1<>  A  15.  Set  tlie 
dividers  to  anv  convenient  spaee.  HIM!  step  oil'  tlie 
oblique  line,  as  A  C,  eight  divisions,  as  shown  by  a  I 
cd,  etc.  From  the  last  of  the  points,  //,  thus  obtained, 
draw  a  line  to  the  end  of  the  given  line,  as  shown  l>v 
h  h*.  Parallel  to  this  line  draw  other  lines,  from  each 
of  the  other  points  to  the  given  line.  The  divisions 


K 


Fig.  1S9.—TO  Bisect  a  Straight  Line  by  the  Use  of  a   Triangle. 

thus  obtained,  indicated  in  the  engraving  by  a"  V  c4, 
etc.,  will  be  the  desired  spaces  in  the  given  line.  It  is 
evident  by  this  rule  that  it  is  immaterial,  except  as  a 
matter  of  convenience,  to  what  space  the  dividers  are 
set.  The  object  of  the  second  oblique  line  in  the  en- 
graving is  to  illustrate  this.  Upon  A  C  the  dividers 
were  set  so  as  to  produce  spaces  shorter  than  those  re- 


g"     b*     c°      d*      e*      /»      „» 


Fig.  340.— To  Divide  a  Given  Straight  Line  into  Any  Number 
of  Equal  Parts 

quired  in  the  given  line  A  B,  while  in  A  D  the  spaces 
were  made  longer  than  those  required  in  the  given  line. 
By  connecting  the  last  point  of  either  line  with  the 
point  B,  as  shown  by  the  lines  h  h'  and  h1  h',  and  draw- 
ing lines  from  the  points  in  each  line  parallel  to  these 
lines  respectively,  it  will  be  seen  that  the  same  divi- 
sions are  obtained  from  either  oblique  line. 

11.  To  Divide  a  Straight  Line  Into  Any  Number  of 
Equal  Parts  by  Means  of  a  Scale.— It  may  be  more  con- 
venient to  transfer  the  length  of  a  given  line  to  a  slip  of 


Geometrical    I 'ml, I  nun. 


3'J 


paper,  rind  by  laving  the  paper  across  a  scale,  as  shown 
in  Fig.  141,  mark  the  required  dimensions  upon  it, 
and  afterward  transfer  them  to  a  given  line,  than  to 
divide  the  line  itself  by  one  of  the  methods  explained 
for  that  purpose.  It  mav  also  occur  that  it  is  desirable 
to  divide  lines  of  different  lengths  into  the  same  num- 
ber of  equal  parts,  or  the  same  lengths  of  lines  into 
different  numbers  of  equal  parts.  Such  a  scale  as  is 
shown  in  Fig.  141  is  adapted  to  all  of  these  purposes. 
The  scale  may  be  ruled  upon  a  piece  of  paper  or  upon 
a  sheet  of  metal,  as  is  preferred.  The  lines  may  be  all 
of  one  color,  or  two  or  more  colors  may  be  alternated. 
in  order  to  facilitate  counting  the  lines  or  following 
them  by  the  eye  across  the  sheet.  In  size,  the  scale 
should  be  adapted  to  the  special  purposes  for  which  it 
is  intended  to  be  used.  By  the  contrast  of  two  colors 


Fig.  HI.— To  Divide  a  Straight  Lin*  into  Any  Number  of 
Equal  Parts  by  Means  of  a  Scale. 


in  ruling  the  lines,  one  scale  may  be  adapted  to  both 
coarse  and  fine  work.  For  instance,  if  the  lines  are 
ruled  a  quarter  of  an  inch  apart,  in  colors  alternating 
red  and  blue,  in  fine  work  all  the  lines  in  a  given  space 
may  be  used,  while  in  large  work,  in  which  the  di- 
mensions are  not  required  to  be  so  small,  either  all  the 
red  or  all  the  blue  lines  may  be  used,  to  the  exclusion 
of  those  of  the  other  color.  Let  it  be  required  to  di- 
vide the  line  A  B  in  Fig.  141  into  thirty  equal1  parts. 
Transfer  the  length  A  B  to  one  edge  of  a  slip  of  paper, 
as  shown  by  A1  B1,  and  placing  A1  against  the  first  line 
of  the  scale,  carry  B'  to  the  thirtieth  line.  Then  mark 
divisions  upon  the  edge  of  the  strip  of  paper  opposite 
each  of  the  several  lines  it  crosses,  as  shown.  Let  it 
be  required  to  divide  the  same  length  A  B  into  fifteen 
equal  parts  by  the  scale.  Transfer  the  length  A  B  to 


a  straight  strip  of  paper,  as  before.  PlacJ  A'  against 
the  first  line  and  carry  B2  against  the  fifteenth  line, 
as  shown.  Then  mark  divisions  upon  the  edtre  of  the 
paper  opposite  each  line  of  the  scale,  as  shown. 

12.  To  Divide  a  Given  Angle  into  Two  Equal  Parts.— 

In  Fig.  142,  let  A  0  B  represent  any  angle  which  it  is 


Fig.  US.— To  Bisect  a  Given  Angle. 

required  to  bisect.  From  the  vertex,  or  point  C,  as 
center,  with  any  convenient  raditis,  strike  the  arc  D  E, 
cutting  the  two  sides  of  the  angle.  From  1)  and  E  as 
centers,  with  any  radius  greater  than  one-half  the 
length  of  the  arc  D  E,  strike  short  arcs  intersecting  at 
G,  as  shown.  Through  the  point  of  intersection,  G, 
draw  a  line  to  the  vertex  of  the  angle,  as  shown  by 
F  C.  Then  F  C  will  divide  the  angle  into  two  equal 
parts. 

13.  To  Trisect  an  Angle.— No  strictly  geometrical 
method  of  solving  this  problem  has  ever  been  discov- 
ered. The  following  method,  partly  geometrical  and 
partly  mechanical,  is,  however,  perfectly  accurate  and 
can  be  used  to  advantage  whenever  it  becomes  neces- 
sary to  find  an  exact  one-third  or  two-thirds  of  an 
angle : 

Let  ABC,  Fig.   143,  be  the  angle  of  which  it  is 


Fig.  143.— To  Trisect  a  Given  Angle. 

required  to  find  one-third.  Extend  one  of  its  sides 
beyond  the  vertex  indefinitely,  as  shown  by  B  E,  and 
upon  this  line  from  B  as  center  with  any  convenient 
radius  describe  a  semicircle  A  C  D,  cutting  both  sides 
of  the  angle.  Place  a  straight  edge  firmly  against  the 


40 


Tlie  Xew  Metal    Worker 


Tiook. 


extended  side  as  at  F,  and  a  pin  at  the  point  C.  On 
another  straight  edge  (Ci)  having  a  perfect  corner  at  K, 
set  oil'  from  one  end  a  distance  equal  to  the  radius 
of  the  semicircle  as  shown  by  point  x\  and  placing 
this  straight  edge,  with  the  end  upon  which  the  radius 
was  set  oil',  against  the  other  straight  edge  (F)  and  its 
edge  near  the  other  end,  against  the  pin  at  the  point  C, 
all  as  shown,  slide  it  along  until  the  mark  x  comes  to 
the  semicircle  establishing  the  point  1).  Draw  the  line 
I)  B,  then  the  angle  D  E  B  will  be  one-third  of  the 
angle  ABC,  and  C  D  B  will  be  two-thirds  of  it. 

14.  To  Find  the  Center  from  which  a  Given  Arc 

is  Struck.— In  Fig.  U4,  let  A  B  C  represent  the  given 


Fig.  144.— To  Find  the  Center 
from  which  a  Given  Arc  is 
Struck. 


Fig.  146 — The  Chord  and  Right 
of  a  Segment  of  a  Circle  Being 
Given,  to  find  the  Center  from 
which  the  Arc  may  be  Struck. 


are,,  the  center  from  which  it  was  struck  being  un- 
known and  to  be  found.  From  any  point  near  the 
middle  of  the  arc,  as  B,  with  any  convenient  radius, 
strike  the  arc  F  G,  as  shown.  Then  from  the  points 
A  and  C,  with  the  same  radius,  strike  the  intersecting 
arcs  I  II  and  E  D.  Through  the  points  of  intersec- 
tion draw  the  lines  K  M  and  L  M,  which  will  meet  in 
M.  Then  M  is  the  center  from  which  the  given  arc 
was  struck.  Instead  of  the  points  A  and  C  being 
taken  at  the  extremities  of  the  arc,  which  would  be 
quite  inconvenient  in  the  case  of  a  long  arc,  these, 
points  may  be  located  in  any  part  of  the  arc  which  is 
most,  convenient.  The  greater  the  distance  between 
A  and  B  and  15  and  C,  the  greater  will  be  the  accu- 
racv  of  succeeding  operations.  The  essential  feature 
of  this  rule  is  to  strike  an  arc  from  the  middle  one  of 
the  points,  and  then  strike;  intersecting  ares  from  the 
other  two  points,  using  tiie  same  radius.  It  is  not 
necessary  that  the  distance  from  A  to  B  and  from  15 
to  C  shall  be  exact! v  the  same. 


15.  To  Find  the  Center  from  which  a  Given  Arc  is 
Struck  by  the  Use  of  the  Square.— In  Kig.  u.v,  let  A 

B  G  be  the  given  arc.  Kstablisli  the  point  B  at  pleas 
nre  and  draw  two  chords,  as  shown  l>v  A  15  and  B  C. 
Misect  these  chords,  obtaining  the  points  E  and  D. 
Place  the  square  against  the  chord  15  C,  as  shown  in 
the  engraving,  bringing  the  heel  against,  the  midd:- 
point,  D,  and  scribe  along  the  Made  indefinitely. 
Then  place  the  square,  as  shown  by  the  dotted  lines, 
with  the  heid  against  the  middle  point,  E,  of  the 
second  chord,  and  in  like  manner  scribe  along  the 
blade,  cutting  the  first  line  in  the  point  F.  ThenF' 
will  be  the  center  of  the  circle,  of  which  tue  are  A  15  C 


Fig.  145.— To   Find  the    Center  from  which  a  Given  Are.  is 
Struck,  by  the  Use  of  the  Square. 


is  a  part.  This  rule  will  be  found  very  convenient  for 
use  in  all  cases  where  the  radius  is  less  than  24  inches 
in  length. 

16.  The  Chord  and  Right  of  a  Segment  of  a  Circle 
being  Given,  to  Find  the  Center  from  which  the  Arc 

may  be  Struck.— In  Fig.  14-C.,  let  A  B  represent  the 
chord  of  a  segment  or  arc  of  a  circle,  and  1)  C  the  rise 
or  hight.  Tt,  is  required  to  find  a  center  from  which 
an  arc,  if  struck,  will  pass  through  the  three  points 
A,  D  and  B.  Draw  A  D  and  B  D.  Bisect  A  D,  as 
shown,  and  prolong  the  line  11  L  indefinitely.  Bisect 
D  B  and  prolong  I  M  until  it  cuts  II  L,  produced  in 
the  point  E.  Then  E,  the  point  of  intersection,  will 
be  the  center  sought.  It  will  be  observed  that  by 
producing  DC,  and  intersecting  it  by  either  H  L  or 


41 


I  M  prolonged,  the  same  point  is  found.  Therefore, 
if  preferred,  ihr  bisecting  of  either  A  Dm-  1.)  15  ma\ 
he  dispensed  with.  A  practical  application  of  this 
rule  occurs  quite  frequently  in  cornice  work,  in  the 
construction  of  window  c:ips  and  other  similar  forum, 
to  lit  frames  already  made.  In  the  conveying  of  or- 
ders from  the  master  builder  or  carpenter  to  the  cor- 
nice worker,  it  is  quite  customar\  to  describe  the  shape 
of  the  head  of  the  frames  which  the  caps  are  to  lit  by 
stating  that  the  width  is,  for  example-.  :;tl  inches,  and 
that  the  rise  is  4  inches.  To  draw  the  shape  thus  de- 
scribed, proceed  as  follows:  Set  oil'  A  1>  equal  to  :'>('> 
inches,  from  the  center  of  which  erect  a  perpendicular, 
D  C,  which  make  equal  to  4  inches.  Continue  D  C  in 
the  direction  of  K  indeiinitely.  Draw  A  1),  which 
bisect,  as  shown,  and  draw  11  L,  producing  it  until  it- 
cuts  D  C  prolonged,  in  the  point  E.  Then  with  E  as 
center  and  K  ])  as  radius,  strike  the  arc  A  D  B. 

17.  To  Strike  an  Arc  of  a  Circle  by  a  Triangular 
Guide,  the  Chord  and  Hight  Being  Given.— In  Fig.  147,  let 


Fig.  14?.— To  Strike  an  Arc  of  a  Circle  by  a  Triangular 

A  D  be  the  given  chord  and  B  F  the  given  hight.  The 
iirst  step  is  to  determine  the  shape  and  size  of  the  tri- 
angular guide.  Connect  A  and  F,  as  shown.  From 
F,  parallel  to  the  given  chord  A  D,  draw  F  (1,  making 
it  in  length  equal  to  A  F,  or  longer.  Then  A  F  (!.  as 
shown  in  the  engraving,  is  the  angle  of  the  triangular 
guide  to  be  used.  Construct  the  guide  of  any  suitable 
material,  making  the  angle  of  two  of  its  sides  equal  to 
the  angle  A  F  G.  Drive  pins  at  the  points  A,  F  and 
D.  Place  the  guide -as  shown.  Put  a  pencil  at  the 
point  F.  Shift  the  guide  in  such  a  manner  that  the 
pend!  will  move  toward  A,  keeping  the  Lruide  at  all 
times  against  the  pins  A  and  F.  Then  reversing,  shift 
the  guide  so  that  the  pencil  at  the  point  F  will  move 
toward  I),  keeping  the  guide  during  this  operation 
atrainst  the  pins  F  and  D.  By  this  means  the  pencil 
will  be  made  to  describe  the  arc  A  F  D.  It  may  be 
interesting  to  know  that  if  the  angle  F  of  the  triangu- 
lar guide  be  made  a  right  angle,  the  .arc  described  by 
it  will  be  a  semicircle.  By  these  means,  then,  a  steel 
square- may  be  used  in  drawing  circles,  as  illustrated 
in  Fig.  148,  the  pins  being  placed  at  A,  B  and  C. 


18.  To  Draw  a  Circle  Through  any  Three  Given 
Points  not  in  a  Straight  Line.— In  Kig.  im,  let  A.  Hand 

K  lie  any  three  given  points  not  in  a  straight  Hue, 
through 'which  it.  is  required  to  draw  a  circle.  Con- 
nect the  given  points  by  drawing  the  lines  A  I)  and  1) 
K.  Bisect  the  line  A  1)  by  F  C,  drawn  perpendicular 
to  it,  as  shown.  Also  bisect  D  K  by  the  line  (!  C,  as 
shown.  Then  the  point  C,  at  which  these  lines  meet, 
is  the  center  of  the  required  circle. 

19.  To  Erect  a  Perpendicular  to  an  Arc  of  a  Circle, 
without  having  Recourse  to  the  Center.— In  Fig.  15o,  let 
A  D  B  be  the  arc  of  a  circle  to  which  it  is  required  to 
erect  a  perpendicular.     With  A  as  center,  and   with 
any  radius  greater  than   half   the  length  of  the  given 
arc,  describe  the   arc   x  x,  and    with   B  as  center,  and 
with  the  same  radius,  describe  the  arc  y  y,  intersecting 


Fig.  IfS. — To  Describe  a  Semicircle  with  a  Steel  Square. 

the  arc  first  struck,  as  shown.  Through  the  points  of 
intersection  draw  the  line  F  E.  Then  F  E  will  be 
perpendicular  to  the  arc,  and  if  sufficiently  produced 
will  reach  the  center  from  which  the  arc  A  B  is  drawn. 

20.  To  Draw  a  Tangent  to  a  Circle  or  Arc  of  a 
Circle  at  a  given  Point  without  having  Recourse  to  the 
Center.— In  Fig.  151,  let  A  D  B  be  the  arc  of  a  circle, 
to  which  a  tangent    is  to  be  drawn  at  the    point  D. 
With  D  as   center,  and,  with   any  convenient   radius, 
describe   the  arc   A   F  B,  cutting  the  given  arc  in  the 
points  A  and  B.      Join  the  points  A  and  B,  as  shown. 
Through   D  draw   a  straight   line   parallel   to  A  B,  as 
shown  by  I1]  II,  then  E  II  will  be  the  required  tangent. 

21.  To  Ascertain  the  Circumference  of  a  Given  Circle. 
— In  Fig.  152,  let  A  D  B  C  be  the  circle,  eqaal  to  the 
circumference  of  which  it  is  required  to  draw  a  straight 
line.      Draw    any   tw:>    diameters    at    right    angles,    as 
shown  by  A  B  and  1)  C.      Divide  one  of  the  four  ai'cs, 
as,  for  instance,  D  1!.  into  eleven  equal  parts,  as  shown. 


The  New   Mi'tal    MV/'vy    l\iU<-,-n    Hook. 


I'Yom  !',  tlie  second  of  these  divisions  from  the  point 
P>.  let  fall  a.  perpendicular  to  A  1>,  as  shown  l>y  '•'  I''. 
To  three  times  the  diameter  of  the  circle  (A  J5  or  DC) 
add  the  length  !)  F,  and  the  result  will  he  a  very  c.h>se 
approximation  to  the  length  of  the  circumference.  This 
rule,  upon  ;\.  diameter  of  1  foot,  gives  a  length  of  about 
T3Tths  of  an  inch  in  excess  of  the  actual  length  of  the 
circumference. 

22.  To  Draw  a  Straight  Line  Equal  in  Length  to  the 
Circumference  of  any  Circle  or  of  any  Part  of  a  Circle  - 


dividers  arc  placed  upon  the  line,  no  perceptible  curve 
shall  exist  between  them,  and,  beginning  at  one  end 
of  the  curve,  step  to  the  oilier  end  of  the  same,  or  so 
near  the  end  tliat  the  remaining  space  shall  be  less  than 
that  between  the  points  of  the  dividers,  then  beginning 
at  the  end  of  any  straight  line  step  oil'  upon  it  the  same 
number  of  spaces,  after  which  add  to  them  the  remain- 
ing small  space  of  the  curve  hv  measurement  with  the 
dividers.  This  will  be  found  the  quickest  and  most 
accurate  of  any  method  for  the  pattern  cutters'  use. 


Fig.  149,— To  Draw  a  Circle  Through 
any  Three  Given  Points  Not  in  a 
Straight  Line. 


Fig.  150. — To  Erect  a  Perpendicular  to 
an  Arc  of  a  Circle. 


Fig.  151.— To  Draw  a   Tangent   to  a  Circle 
or  Arc. 


./ft 


Fig.  IBS. — To  Ascertain  the  Circumfer- 
ence of  a  Given  Circle. 


Fig.  15S. — To  Inscribe  an  Equilateral 
Triangle  within  a  Given  Circle. 


Fig.  154. — To  Inscribe  a  Square  ivithin 
a  Given  Circle. 


Various  approximate  rules,  similar  to  the  one  given  in 
the  problem  above,  for  performing  these  operations  are 
known  and  sometimes  used  among  workmen,  but  can- 
not be  recommended  here  because  in  using  them  con- 
siderable time  and  trouble  is  required  to  obtain  a  result 
which  is  not  accurate  when  obtained,  thus  rendering 
such  methods  impracticable.  The  simplest  and  most 
accurate  method  for  obtaining  the  length  of  any  curved 
line  is  as  follows :  Take  between  the  points  of  the 
dividers  a  space  so  small  that  when  the  points  of  the 


The  most  common  rules  in  use  for  the  construc- 
tion of  polygons,  whether  drawn  within  circles  or 
erected  upon  given  sides,  are  those  which  einplov  the 
straight-edge  and  compasses  only.  Other  instruments 
may  also  be  employed  to  great  advantage,  as  will  be 
shown  further  on,  leaving  the  student  to  decide  which 
method  is  the  most  suited  to  any  case  he  may  have  in 
hand.  Accordingly,  the  construction  of  polygons  will 
be  treated  under  three  different  heads  arranged  accord- 
ing to  the  tools  employed. 


•I  I! 


THE  CONSTRUCTION  OF  REGULAR  POLYGONS. 


I  —BY  THE  USE  OF  COMPASSES  AND  STRAIGHT-EDGE. 

23.  To  Inscribe  an  Equilateral  Triangle  within  a 
Given  Circle.— In  Fig.   Ij;;,  let  A  B  D  be  any  given 
circle  within    which    an   equilateral    triangle  is  to  be 
dr:iwn.      From  any  point  in   the   circumference,  as  K, 
with  a  radius  equal  to  the  radius  of  the  circle,  describe 
the  arc  D  C  B,  cutting  the  given  circle  in  iho    points 
1)  and  B.      Draw  the   line  D  B,  which  will  be  one  side 
of    the  required  triangle.      From  D  or  B  as  center,  and 
with  1)  B  as  radius,  cut  the  circumference  of  the  given 
cirele,  as   shown    at    A.      Draw  A  B   and   A  D,  which 
will  complete  the  iignre. 

24.  To  Inscribe  a  Square  within  a  Given  Circle.— 

In  Fig.  154,  let  A  C  B  1)  bo   any  given   circle   within 
which  it  is  required  to  draw  a  square.      Draw  any  two 


Circle.— In  Fig.  i:.i;,  let  A  B  D  K  F  G  be  any  given 
circle  within  which  a  hexagon  is  to  be  drawn.  From 
any  point  in  the  circumference  of  the  circle,  as  at  A, 
with  a  radius  equal  to  the  radius  of  the  circle,  de- 
scribe the  arc  C  B,  cutting  the  circumference  of  the 
circle  in  the  point  B.  Connect  the  points  A  and  B. 
Then  A  B  will  be  one  side  of  the  hexagon.  With  the 
dividers  set  to  the  distance  A  B,  step  off  in  the  cir- 
cumference of  the  circle  the  points  G,  F,  E  and  D. 
Draw  the  connecting  lines  A  G,  (I  F,  F  E,  E  D  and 
1)  B,  thus  completing  the  figure.  By  inspection  of 
this  ligure  it  will  be  noticed  that  the  radius  of  a  circle 
is  equal  to  one  side  of  the  regular  hexagon  which  may 
be  inscribed  within  it.  Therefore  set  the  dividers  to 
the  radius  of  a  circle  and  step  around  the  circumfer- 
ence, connecting  the  points  thus  obtained. 


Fig.  155  —To  Inscribe  a  Regular  Penta- 
gon within  a  Given  Circle. 


Fig.  15fi.—To  Inscribe  a  Regular  Hexa- 
gon within  a  Given  Circle. 


Fig.  157.— To  Inscribe  a  Regular  Hepta- 
gon \uithin  a  Given  Circle. 


diameters  at  right  angles  with  each  other,  as  C  D  and 
A  B.  Join  the  points  C  B,  B  D,  D  A  and  A  C,  which 
will  complete  the  required  figure. 

25.  To  Inscribe  a  Regular  Pentagon  within  a  Given 
Circle. — In  Fig.    loo,   A  D  B  G  represents  a  circle  in 
which  it  is  required  to  draw  a  regular  pentagon.    Draw 
any  two  diameters  at    right  angles  to  each  other,   as 
A  B  and  D  C.      Bisect  the  radius  A  II,  as  shown  at  E. 
With  E  as  center  and   E  D  as  radius   strike  the   arc  D 
F,  and  with  the  chord  D  F  as  radius,  from  D  as  center, 
strike  the  arc  F  G,  cutting  the  circumference  of  the 
given  circle  at  the  point  G.      Draw  D  G,  which  will 
equal    one    side    of    the    required    figure.      With    the 
dividers  set  equal  to  D  G,  step  off  the  spaces  in  the 
circumference  of   the   circle,  as  shown  by  the  points 
I  K  and  L.      Draw  D  I,  I  K,  K  L  and  L  G,  thus  com- 
pleting the  figure. 

26.  To  Inscribe  a  Regular  Hexagon  within  a  Given 


2f.  To  Inscribe  a  Regular  Heptagon  within  a  Given 
Circle.— In  Fig.  157,  let  F  A  G  B  II  I  K  L  D  be  the 

given  circle.  From  any  point,  A,  in  the  circumfer- 
ence, with  a  radius  equal  to  the  radius  of  the  circle, 
describe  the  arc  BCD,  cutting  the  circumference  of 
the  circle  in  the  points  B  and  D.  Draw  the  chord  B  D. 
Bisect  the  chord  B  D,  as  shown  at  E.  With  D  as 
center,  and  with  D  E  as  radius,  strike  the  arc  E  F, 
cutting  the  circumference  in  the  point  F.  Draw  D  F, 
which  will  be  one  side  of  the  heptagon.  With  the 
dividers  set  to  the  distance  D  F,  set  off  in  the  circum- 
ference of  the  circle  the  points  G  II  I  K  and  L,  and 
draw  the  connecting  lines  F  G,  G  II,  II  I,  I  K,  K  L 
and  L  D,  thus  completing  the  figure. 

28.  To  Inscribe  a  Regular  Octagon  within  a  Given 
Circle.— In  Fig.  158,  let  B  I  D  F  A  G  E  II  be  the 
given  circle  within  which  an  octagon  is  to  be  drawn. 
Draw  any  two  diameters  at  right  angles  to  each  other, 


44 


The  New  Metal   Worker  Pattern  Book. 


as  15  A  ami  1)  E.  Draw  the  chords  D  A  and  A  E. 
Bisect  13  A,  as  shown,  and  draw  L  II.  Bisect  A  K 
and  draw  K  1.  Then  connect  llie  several  points  in  the 
circumference  thus  obtained  liv  drawing  the  lines  I)  I, 
I  11,  B  II,  II  E,  K  G,  G  A,  A  F  and  F  D,  which  will 
complete  the  figure. 

29.  To  Inscribe  a  Regular  Nonagon  within  a  Given 
Circle.—  ! n  Fig.  l.V.»,  let  M  K  F  E  l>e  the  given  circle. 
Draw  nnv  two  radii  at  right  angles  to^ach  other,  as  1> 
(}  and  A  C,  and  draw  the  chord  B  A.  From  A  as 
center,  and  with  a  radius  equal  to  one-half  the  chord 
A  B,  as  shown  l>y  A  D,  strike  the  arc  D  E,  cutting  the 
circumference  of  the  circle  at  the  point  E.  Draw  A  E, 
which  will  he  one  side  of  the  nonagon.  Set  the  di- 
viders to  the  distance  A  E  and  step  off  the  points  M, 
II,  K,  G,  I,  F  and  L,  and  draw  the  connecting  lines, 
as  shown,  thus  completing  the  figure. 


Circle.— In  Fig.  101,  let  B  D  A  L  ho  any  given  circle 
in  which  a  regular  iignre  of  eleven  sides  is  to  be 
drawn.  Draw  any  diameler,  as  J>  A.  and  draw  a 
radius,  as  D  G,  at  right  angles  to  B  A.  Bisect  C  A, 
thus  obtaining  the  point  E.  From  E  as  center,  and 
with  K  D  as  radius,  describe  the  arc  D  F,  cutting  P>  A 
in  the  point  F.  With  D  as  center,  and  1)  F  as  radius, 
describe  the  arc  F  G,  cutting  the  circumference  in  the 
point.  G.  Draw  the  chord  G  D  and  bisect  it,  as  shown 
by  II  C,  thus  obtaining  the  point  K.  From  D  as 
center,  and  with  D  K  as  radius,  cut  the  circumference 
in  the  point  I.  Draw  I  D.  Then  I  D  will  be  equal 
to  one  side  of  the  required  figure.  Set  the  dividers  to 
this  space  and  step  off  the  points  in  the  circumference, 
as  shown  by  N,  E,  S,  M,  P,  L,  0,  T  and  V,  and  draw 
the  connecting  chords,  as  shown,  thus  completing  the 
figure. 


8     M 


Fig.  1~>8. — To  Inscribe  a  Regular  Octa- 
gon within  a  Oin-n  Circle. 


Fig.  159. — To  Inscribe  a  Regular  Nuna- 
gon  within  a  Given  Circle. 


Fig.  160. — To  Inscribe  a  Regular  Deca- 
gon within  a  Given  Circle. 


30.  To  Inscribe  a  Regular  Decagon  within  a  Given 

Circle.— In  Fig.  100,  let  D  B  E  A  be  any  given  circle 
in  which  a  decagon  is  to  be  drawn.  Draw  any  two 
diameters  through  the  circle  at  right  angles  to  each 
other,  as  shown  by  B  A  and  D  E.  Bisect  B  C,  as 
shown  at-^F,  and  draw  F  D.  With  F  as  center,  and 
F  D  as  radius,  describe  the  are  D  G,  cutting  B  A  in  the 
point  G.  Draw  the  chord  D  G.  "With  D  as  center, 
and  D  G  as  radius,  strike  the  are  G  II,  cutting  the  cir- 
cumference in  the  point  II.  Connect  D  and  ]I,  as 
shown.  Bisect  D  II  and  draw  the  line  C  It,  cutting 
the  circumference  in  the  point  I.  Draw  the  lines  II  I 
and  I  D,  which  will  then  be  two  sides  of  the  required 
figure.  Set  the  dividers  to  the  distance  II  I  and  space 
oil'  the  circumference  of  the  circle,  as  shown,  and  draw 
the  connecting  lines  D  K,  K  M,  M  L,  L  P,  P  E,  E  N, 
N  O  and.  O  II,  thus  completing  the  ligniv. 

31.  To  Inscribe  a  Regular  Undecagon  within  a  Given 


32.  To  Inscribe  a  Regular  Dodecagon  within  a  Given 
Circle. — In  Fig.  102,  let  M  F  A  I  be  any  given   circle 
in    which    a    dodecagon    is    to  be  drawn.      From  anv 
point  in  the  circumference,  as  A.  with  a  radius  equal 
to  the  radius  of   the   circle,  describe  the  arc  C  B,  cut- 
ting   the    circumference    in   the    point    B.      Draw    the 
chord  A  P>,  which  bisect  as  shown,  and  draw  the  line 
0  C,  cutting  the  circumference  in  the  point  D.      Draw 
A  D,  which  will  then  be  one  side  of  the  given  figure. 
With  the  dividers  set  to  this  space    step  off  in  the  cir- 
cumference the  points  B,  I,  X,  II,  M,  G,  L,  F,  K  and 
F,  and  draw  the  several   chords,  as  shown,  thus  com- 
pleting the  Iignre. 

33.  General  Rule  for  Inscribing  any  Regular  Polygon 

in  a  Given  Circle. — Through  the  given  circle  draw  any 
diameter.  At  right  angles  to  this  diameter  draw  a 
radius.  Divide  that  radius  into  four  equal  parts,  ami 
prolong  it  outside  the  circle  to  a  distance  equal  to  three 


< i,  nun  triful    Problt  in*. 


45 


of  those  parts.  Divide  the  diameter  of  the  circle  into 
the  same,  number  of  equal  parts  as  the  polygon  is  to 
have  sides.  Then  1  mm  1  lie  end  of  the  radius  prolonged, 
as  above  described,  through  the  second  division  in  the 
diameter,  draw  a  line  cutting  the  ci  renni  ferenee.  Con- 
neet  this  point,  in  the  circumference  and  the  nearest 
end  of  the  diameter.  The  line  thus  drawn  will  lie  one 
side  of  the  required  ligure.  Set  the  dividers  to  this 
space  and  step  oil  on  the  circumference  of  the  eirclo 


outside  ihe  circle  to  the  extent,  of  three  of  those'  parts, 
as  sho\vn  1^\  abe,  thus  obtaining  the  point  <:.  From 

r,  through  the  second  division  in  the  diameter,  draw 
the  line  c  II,  cutting  tlie  circumference  in  the  point 
II.  Connect  II  and  K.  Then  JI  K  will  be  one  side 
of  the  required  ligure.  Set  the  dividers  to  the  dis- 
tance 11  Hand  step  oil  the  circumference,  as  shown, 
thus  obtaining  ihe  points  for  the  other  sides,  and  dr»w 
the  connecting  arcs,  all  us  illustrated  in  the  ligurc. 


f,V,   161.— To  Iiixciibe   a.  Regular   Un- 
decagnn  within  a  Given  Circle. 


Fig.  162.— To  Inscribe  a  Regular  Dodec- 
agon within  a  Given  Circle. 


Fig.  ItiJ.—To  Inscribe  a  Regular  Undec- 
agun  within  a  Given  Circle  by  Ihe 
General  Rule. 


Q 


Fig.  M./.  —  Ujiun  a  (liom  Side  lo 
Construct  an  Equilateral  Tri- 
angle. 


Fig.  Via. — To  Construct  a  Triangle,  Ihe 
Length  of  the  Three  Sides  being 
(liren. 


Fig.    166. — I 'pan    a    Givm    Side     to   draw    n 
Regular  Pentagon, 


the  remaining  number  of  sides  and  draw  connecting 
lines,  which  will  complete  the  ligurc. 

34.  To  Inscribe  a  Regular  Polygon  of  Eleven  Sides 
(Undecagon)  within  a  Given  Circle  by  the  General  Rule. 

—Through  the  given  circle,  E  I)  F  G  in  Fig.  1«3, 
draw  any  diameter,  as  K  K.  which  divide  into  the 
same  number  of  equal  parts  as  the  ligurc  is  to  have 
sides,  as  shown  by  the  small  figures.  At  right,  angles 
to  the  diameter  just  drawn  draw  the  radius  D  K,  which 
divide  into  four  equal  parts.  Prolong  the  radius  D  K 


35.  Upon  a  Given  Side  to  Construct  an  Equilateral 
Triangle.— In  Fig.  lf.4,  let  A  B   represent   the   length 
of  the  given  side.      Draw   any  line,  as  C  D,  making  it 
equal  to  A  B.      Take  the  length  A  B    in   the  dividers, 
and  placing  one  foot    upon    the    point  C,  describe   the 
are    K  F.      Then    from    D    as    center,    with    the    same 
radius,  describe  the  are  (i  II,   intersecting  the  first    arc- 
in    the  point   K.      Draw  K  C  and  K  D.      Then  C  D  K 
will  be  the  required  triangle. 

36.  To  Construct  a  Triangle,  the  Length  of  the  Three 


Tlte   Xew   Metal    \\~orker  Patlf. 


i 


Sides  being  Given.— In  Fig.  165,  let  A  B.  c  D  and  K 
F  be  the  given  sides  from  which  it  is  required  to  con- 
struct a  triangle.  Draw  any  straight  line,  G  II,  mak- 
ing it  in  length  equal  to  one  of  the  sides,  E  F.  Take 
the  length  of  one  of  the  other  sides,  as  A  B,  in  the 
compasses,  and  from  one  end  of  the  line  just  drawn, 
as  G,  for  center  describe  an  arc,  as  indicated  by  L 
M.  Then  set  the  compasses  to  the  length  of  third 
side,  C  1),  and  from  the  opposite  end  of  the  line 
first  drawn,  H,  describe  a  second  arc,  as  I  K,  intersect- 
ing the  first  in  the  point  0.  Connect  0  G  and  0  H. 
Then  0  G  II  will  be  the  required  triangle. 

37.  Upon  a  Given  Side  to  Draw  a  Regular  Pentagon. 
— In  Fig.  166,  let  A  B  represent  the  given  side  upon 
which  a  regular  pentagon  is  to  be  constructed.  With 
B  as  center  and  B  A  as  radius,  draw  the  semicircle 


of  the  circle,  obtaining  the  points  M 
ami  L.  Draw  A  M,  M  L  and  L  D,  which  will  com- 
plete the  figure. 

38.  Upon  a  Given  Side  to  Draw  a  Regular  Hexagon. 
— In  Fig.  167,  let  A  B  be  the  given  side  upon  which 
a  regular  hexagon  is  to  be  erected.  From  A  as  center, 
and  with  A  B  as  radius,  describe  the  arc  B  C.  From 
B  as  center,  and  with  the  same  radius,  describe  the 
arc  A  C,  intersecting  the  first  arc  in  the  point  C.  C 
will  then  be  the  center  of  the  circle  which  will  cir- 
cumscribe the  required  hexagon.  With  C  as  cent  IT. 
and  C  B  as  radius,  strike  the  circle,  as  shown.  Set 
the  dividers  to  the  space  A  B  and  step  off  the  circum- 
ference, as  shown,  obtaining  the  points  E,  G,  F  and  D. 
Draw  the  chords  A  E,  E  G,  G  F,  F  D  and  D  B,  thus 
completing  the  required  figure. 


Fig.  167.— Upon  a   (liven  Side  to  Draw 
a  Regular  Hexagon. 


Fig.    16$.—l'pon  a  Giivji   Side  to  Draw 
a  lieyular  Heptagon. 


Fig.  KO.—Upon  a  Given  Side  to    llmw 
a  Regular    Octagon. 


A  D  E.  Produce  A  B  to  E.  Bisect  the  given  side 
A  J3,  as  shown  at  the  point  F,  and  erect  a  perpendic- 
ular, as  shown  by  F  C.  Also  erect  a  perpendicular 
at  the  point  B,  as  shown  by  G  H.  With  B  as  center, 
and  F  B  as  radius,  strike  the  arc  F  G,  cutting  the  per- 
pendicular H  G  in  the  point  G.  Draw  G  E.  With 
G  as  center,  and  G  E  as  radius,  strike  the  arc  E  II, 
cutting  the  perpendicular  in  the  point  H.  With  E  as 
center,  and  E  H  as  radius,  strike  the  arc  II  D,  cutting 
the  semicircle  A  D  E  in  the  point  D.  Draw  D  B, 
which  will  be  the  second  side  of  the  pentagon.  Bisect 
D  B,  as  shown,  at  the  point  K,  and  erect  a  perpendic- 
ular, which  produce  until  it  intersects  the  perpendic- 
ular F  C,  erected  upon  the  center  of  the  given  side  in 
point  F.  Then  C  is  the  center  of  the  circle  which 
circumscribes  the  required  pentagon.  From  C  as  cen- 
ter, and  with  C  B  as  radius,  strike  the  circle,  as  shown. 
Set  the  dividers  to  the  distance  A  B  and  step  off  the 


39.  Upon  a  Given  Side  to  Draw  a  Regular  Heptagon. 

— In  Fig.  168,  A  B  represents  the  given  side  upon 
which  a  regular  heptagon  is  to  be  drawn.  From  B  as 
center,  and  with  B  A  as  radius,  strike  the  semicircle 
A  E  D.  Produce  A  B  to  D.  From  A  as  center,  ami 
with  A  B  as  radius,  strike  the  arc  B  F,  cutting  the 
semicircle  in  the  point  F.  Through  F  draw  F  G  per- 
pendicular to  A  B,  which  extend  in  the  direction  of  C. 
From  D  as  center,  and  with  radius  G  F,  cut  the  semi- 
circle in  the  point  E.  Draw  the  line  E  B,  which  is 
another  side  of  the  required  heptagon.  Bisect  E  B,  and 
upon  its  middle  point,  H,  erect  a  perpendicular,  which 
produce  until  it  meets  the  perpendicular  erected  upon 
the  center  of  the  given  side  A  B,  in  the  point  C.  Then 
0  is  the  center  of  the  circle  which  will  circumscribe 
the  required  heptagon.  From  C  as  center,  and  with 
C  B  as  radius,  strike  the  circle.  Set  the  dividers  to 
the  distance  A  B  and  step  off  the  circumference,  as 


Geometrical   Problems. 


47 


shown,  obtaining  the  points  K,  N,  M  and  L.  Draw 
the  connecting  arcs  A  K,  K  N,  N  M,  M  L  and  L-E, 
thus  completing  the  figure. 

40.  Upon  a  Given  Side  to  Draw  a  Regular  Octagon.— 

In  Fig.  10!>,  let  A  1>  represent  the  given  side  upon 
which  a  regular  octagon  is  to  bo  constructed.  Produce 
A  13  indefinitely  in  the  direction  of  D.  From  B  as 
center,  and  with  A  H  as  radius,  describe  the  semicircle 
A  E  D.  At  the  point  B  erect  a  perpendicular  to  A  B, 
as  shown,  cutting  the  circumference  of  the  semicircle 
in  the  point  E.  Bisect  the  arc  E  D,  obtaining  the 
point  F.  Draw  F  15,  which  is  another  side  of  the  re- 
quired octagon.  Bisect  the  two  sides  now  obtained 
and  erect  perpendiculars  to  their  middle  points,  G  and 
H,  which  produce  tmtil  thev  intersect  at  the  point  C. 
(J  "then  is  the  center  of  the  circle  that  will  circumscribe 


which  produce  until  they  intersect  at  the  point  C. 
Then  C  is  the  center  of  the  circle  which  will  circum- 
scribe the  required  nonagon.  From  C  as  center,  and 
with  C  B  as  radius,  strike  the  circle  B  0  P  A.  Set 
the  dividers  to  the  space  A  B  and  step  oil  the  circle, 
as  shown,  obtaining  the  points  N,  P,  M,  E,  0  and  L. 
Draw  the  connecting  chords,  A  N,  N  P,  P  M,  M  R, 
R  0,  0  L  and  L  E,  thus  completing  the  figure. 

42.  Upon  a  Given  Side  to  Draw  a  Regular  Decagon. 
— In  Fig.  171,  A  B  is  the  given  side  upon  which  a 
regular  decagon  is  to  be  drawn.  Produce  A  B  indefi- 
nitely in  the  direction  of  D.  From  B  as  center,  and 
with  B  A  as  radius,  strike  the  semicircle  A  H  D. 
Bisect  the  given  side  A  B,  obtaining  the  point  F. 
Through  the  point  B  draw  the  line  H  B  G,  perpen- 
dicular to  A  B.  From  B  as  center,  and  with  B  F  as 


Fig.  170.— Upon  a  Given  Side  to  Draw 
a  Regular  Nonagon. 


Fig.  171  —  I  'pan  a  Given  Sid<:  to  Draw 
a  Regular  Decagon. 


Fig.  172.— Upon  a  Given  Side  to 
Draw  a  Regtilar  Undecagon. 


the  octagon.  From  C  as  center,  and  with  C  B  as  radius, 
strike  the  circle,  as  shown.  Set  the  dividers  to  the 
space  A  B  and  step  off  the  circumference,  obtaining 
the  points  L,  K,  M,  0  and  N.  Draw  the  connecting 
arcs  A  L,  L  K,  K  M,  M  0,  O  N  and  N"  F,  thus  com- 
pleting the  required  figure. 

41.  Upon  a  Given  Side  to  Draw  a  Regular  Nonagon.— 
In  Fig.  170,  A  B  is  any  given  side  upon  which  it  is 
required  to  draw  a  regular  nonagon.  Produce  A  B  in- 
definitely in  the  direction  of  D.  From  B  as  center, 
and  with  B  A  as  radius,  strike  the  semicircle  A  F  D. 
At  the  point  B  erect  a  perpendicular  to  A  B,  cutting 
the  semicircle  in  the  point  F.  Draw  the  chord  F  D, 
which  bisect.,  obtaining  the  point  G.  From  D  as  cen- 
ter, and  with  D  <!  as  radius,  cut  the  semicircle  in  the 
point  E.  Draw  K  B,  which  will  be  another  side  of  the 
required  figure.  From  the  middle  points  of  the  two 
sides  now  obtained,  as  H  and  K,  erect  perpendiculars, 


radius,  strike  the  arc  F  G,  cutting  the  perpendicular 
H  G  in  the  point  G.  From  G  as  center,  and  with  G 
D  as  radius,  strike  the  arc  D  0,  cutting  the  perpen- 
dicular H  G  in  the  point  0.  From  D  as  center,  and 
with  D  0  as  radius,  strike  the  arc  0  K,  cutting  the 
semicircle  in  the  point  K.  Draw  the  line  K  D,  which 
bisect  with  the  line  B  L,  cutting  the  semicircle  in  the 
point  E.  Then  E  B  will  be  another  side  of  the  deca- 
gon. Upon  the  middle  points,  F  and  M,  of  the 
two  sides  now  obtained  erect  perpendiculars,  which 
produce  until  they  intersect  at  the  point  C.  Then  C 
is  the  center  of  the  circle  which  will  circumscribe  the 
required  decagon.  From  G  as  center,  and  with  C  B 
;is  radius,  strike  the  circle,  as  shown.  Set  the  dividers 
to  the  space  A  B  and  step  off  the  circle,  obtaining  the 
several  points,  I,  N,  S,  V,  R,  T  and  P.  Draw  the 
connecting  lines,  A  I,  I  N,  N  S,  S  V,  V  R,  B  T,  T 
P  and  P  E,  thus  completing  the  figure. 


New   M'-tnl    1 1 \,rker  Pattern  Jlook. 


43.  Upon  a  Given  Side  to  Draw  a  Regular  Undec- 


agon. 
upon 


In     Kig.     IT-,    -V   13   represents   the 
regular    undecan-on    is    to 


'6- 

which    a 


Produce  A  13  indefinitely  in  the  direction  of  D.  From 
13  as  center,  and  with  B  A  as  radius,  draw  the  semi- 
circle A  M  B.  Through  the  point  B,  perpendicular 
to  A  B,  draw  the  line  II  D  indefinitely.  From  13  as 
center,  and  with  B  F  as  radius,  strike  the  arc  F  G, 
cutting  the  perpendicular  II  G  in  the  point  G.  From 
G  as  center,  and  G  D  as  radius,  strike  the  arc  1)  1 1 , 
cutting  the  perpendicular  II  G  in  the  point  H.  With 
D  as  center,  and  D  H  as  radius,  strike  the  arc  H  M, 
cutting  the  semicircle  in  the  point  M.  Draw  M  D, 
which  bisect,  obtaining  the  point  K,  through  which, 
from  B,  draw  the  line  B  K,  and  produce  it  until  it 
cuts  the  semicircle  in  the  point  E.  Then  B  E  will  be 
another  side  of  the  required  figure.  Bisect  the  two 
sides  now  obtained  and  erect  perpendicular  lines,  pro- 


N^ 

Fig.  173.— Upon  a  Given  Side  to  Draw  a  Regular  Dodecagon. 

dueing  them  until  they  intersect,  as  shown  bv  F  C 
and  L  C.  Then  C,  the  point  of  intersection,  is  the 
center  of  the  circle  which  circumscribes  the  undec- 
agon.  From  G  as  center,  and  with  C  A  as  radius, 
strike  the  circle,  as  shown.  Set  the  dividers  to  the 
space  A  B  and  step  off  the  circumference,  obtaining 
the  points  O,  V,  T,  E,  P,  S,  N  and  I.  Draw  the 
chords  A  O,  O  V,  V  T,  T  E,  E  P,  P  S,  S  N,  N  I  and 
I  E,  thus  complcuing  the  figure. 

44.  Upon  a  Given  Side  to  Draw  a  Regular  Dodecagon. 
—In  Fig.  173,  let  A  B  represent  tin;  given  side  upon 
which  a  regular  dodecagon  is  to  be  drawn.  Produce 
A  B  indefinitely  in  the  direction  of  D.  From  13  as 
center,  and  with  B  A  as  radius,  describe  the  semicircle 
A  F  D.  From  D  as  center,  and  with  D  B  as  radius, 
describe  the -arc  13  K,  cutting  the  semicircle  in  the 
point  F.  Draw  F  I),  which  bisect  by  the  line  V  1!, 
cutting  the  semicircle  in  the  point  E.  Then  E  15  is 
another  side  of  the  dodecagon.  From  the  middle 


points  of  the  two  sides  now  obtained,  as  G  and  II,  erect 
perpendiculars,  as  shown,  'jutting  each  other  at  the 
point  C.  This  point  of  intersectr:n,  (J.  then  is  the 
center  of  the  circle  which  will  circumscribe  the  required 
dodecagon.  From  C  as  center,  and  with  C  15  as  radius, 
strike  the  circle,  as  shown.  Set  the  dividers  to  the 
distance  A  13  and  space  oil' the  circumference,  thus  ob- 
taining the  points  L,  P.  M,  S,  N,  K.  O,  K  and  1. 
Draw  the  connecting  lines  L  P,  P  M,  M  S,  S  N,  N  E. 
li  0,  0  K,  K  I  and  I  E,  thus  completing  the  figure. 

45.  General  Rule  by  which  to  Draw  any  Regular 
Polygon,  the  Length  of  a  Side  Being  Given.— With  a 

radius  equal  to  the  given  side  describe  a  semicircle,  the 
circumference  of  which  divide  into  as  many  equal  parts 
as  the  figure  is  to  have  sides.  From  the  center  bv 
which  the  semicircle  was  struck  draw  a  line  to  the 


Fig.  174.— Upon  a  Given  Side  to  Construct  a  Regular  Polygon 
of  Thirteen  Sides  by  the  General  Rule. 

second  division  in  the  circumference.  This  line  will 
be  one  side  of  the  required  figure,  and  one-half  of  tin- 
diameter  of  the  semicircle  will  be  another,  and  the  two 
will  be  in  proper  relationship  to  each  other.  There- 
fore, bisect  each,  and  through  their  centers  erect  per- 
pendiculars, which  produce  until  they  intersect.  The 
point  of  intersection  will  be 'the  center  of  the  circle 
which  will  circumscribe  the  polygon.  Draw  the  circle, 
and  setting  the  dividers  to  the  length  of  one  of  the 
sides  already  found,  step  off  the  circumference,  thus 
obtaining  points  by  which  to  draw  the  remaining  sides 
of  the  figure. 

46.  To  Construct  a  Regular  Polygon  of  Thirteen  Sides 
by  the  General  Rule,  the  Length  of  a  Side  being  Given.— 

In  Fig.  174,  let  A  13  be  the  given  side.  With  13  as 
center,  and  with  13  A  as  radius,  describe  the  semicircle 
A  F  G.  Divide  the  circumference  of  the  semicircle 


Geometrical    Problems. 


into  tliirtec'ii  equal  puns,  as  shown  by  the  small  figures, 
1,  2,  3,  4,  etc.  From  B  draw  a  line  to  the  second 
division  in  the  circumference,  as  shown  by  B  2.  Then 
A  B  and  B  '2  are  two  of  the  sides  of  the  required  figure, 
and  are  in  correct  relationship  to  each  other.  Bisect 
A  B  and  B  2,  as  shown,  and  draw  D  C  and  E  C  through 
their  central  points,  prolonging  them  until  they  inter- 
sect at  the  point  C.  Then  C  is  the  center  of  the  circle 
which  will  circumscribe  the  required  polygon.  Strike 
the  circle,  as  shown.  Set  the  dividers  to  the  space 
A  B,  and  step  off  corresponding  spaces  in  the  circum- 
ference of  the  circle,  as  shown,  and  connect  the  several 
points  so  obtained  by  lines,  thus  completing  the  figure. 

47.  Within  a  Given  Square  to  Draw  a  Regular  Octa- 
gon.—In  Fig.  175,  let  A  D  B  E  be  any  given  square 
within  which  it  is  required  to  draw  an  octagon.  Draw 
the  diagonals  D  E  and  A  B,  intersecting  at  the  point 
C.  From  A,  D,  B  and  E  as  centers,  and  with  radius 
equal  to  one-half  of  one  of  the  diagonals,  as  A  C,  strike 
the  several  arcs  H  N,  G  K,  I  M  and  L  O,  cutting  the 
sides  of  the  square,  as  shown.  Connect  the  points 
thus  obtained  in  the  sides  of  the  square  by  drawing 
the  lines  G  O,  H  I,  K  L  and  M  N,  thus  completing 
the  figure. 

For  general  use  a  very  convenient  scale  may  be 
constructed,  as  shown  in  Fig.  176,  from  which  half 
the  length  of  one  side  of  a  polygon  of  any  number  of 
sides  and  of  any  diameter  in  inches  and  fractions  of 


H 


7|B 


M 


0  N  E 

Pig.  175. — Within  a  Oiven  Square  to  Draw  a  Regular  Octagon. 

inches  may  readily  be  obtained.     Draw  the  vertical 
line  OB  and  divide  it  into  inches  and  parts  of  an  inch. 
From  these   points  of   division  draw  horizontal  lines ; 
from  the  point  O  draw  the  following  lines  and  at  the 
following  angles  from  the  horizontal  line  O  P : 
A  line  at  75°  for  polygons  having  12  sides. 
72°  "        10     " 

"  "          8     " 


A  line  at  00°  for  polygons  having    6  sides. 


54° 

45° 


5     " 
4      " 


The  figures  on  0  B  will  designate  the  radius  of 
the  inscribed  circle  measured  from  O.  The  distance 
from  0  B  on  any  horizontal  line  to  the  oblique  line  de- 


Fig.  176  — tfr.it.le.  for  Constructing  Polygons  of  any  Number 
of  Sides,  the  Diameter  of  the  Inscribed  Circle  Being 
Given  in  Inches.— Half  Full  Size. 


noting  the  required  polygon  will  be  half  the  length  of 
a  side  of  the  polygon  of  the  diameter  indicated  by  the 
figure  at  the  end  of  the  horizontal  line  assumed.  The 
distance  from  O  measured  upon  the  oblique  line  to  the 
assumed  horizontal  line  will  be  the  radius  of  the  cir- 
cumscribed circle. 

In  the  engraving  three  polygons  are  drawn  show- 
ing the  application  of  the  scale. 

H.— BY  THE  USE  OF  THE   T-SQUARE    AND    TRIAN- 
GLES  OR    SET-SQUARES. 

In  the  chapter  upon  terms  and  definitions  under 
the  word  degree  (def.  68)  and  in  some  of  those  immedi- 
ately following  the  dimensions  of  the  circle  are  de- 
scribed and  their  use  explained;  and  in  the  chapter 
upon  Drawing  Tools  and  Materials  (on  page  21)  the  tri- 
angles or  set-squares  in  common  use  are  described  and 
illustrated.  As  all  regular  polygons  depend,  for  their 
construction,  upon  the  equal  division  of  the  circle, 
some  explanation  of  the  application  of  the  foregoing 
will  serve  to  fix  a  few  facts  in  the  mind  of  the  student 
and  thus  prepare  him  for  the  use  of  the  set-square. 


50 


Tin-   \>'/r   Mi'lal    Worker  Pattern  Bwk. 


A  well-known  and  easily  demonstrated  geometrical 
principle  is  that  the  sum  of  the  three  interior  angles  of  a 
triangle  is  equal  to  two  right  angles,  or  in  other  words, 
as  a  right  angle  is  one  of  90  degrees,  if  the  three  angles 
of  any  triangle  be  added  together  their  sum  will  equal 
180  degrees.  Hence,  if  one  of  the  angles  of  a  set- 
square  be  fixed  at  90  degrees  (which  is  done  for  con- 
venience in  drawing  perpendicular  lines)  the  sum  of 
the  two  remaining  angles  must  also  be  90  degrees, 
and,  if  then  the  two  other  angles  be  made  equal,  each 
will  be  45  degrees,  which  is  the  half  of  90  degrees. 
If,  however,  one  of  the  other  angles  is  fixed  at  30 
(one-third  of  90  degrees),  the  remaining  angle  must  be 
60  degrees,  as  30  +  60  =  90. 

By  means,  then,  of  the  45-degree  and  the  30  X 
60-degree  triangles,  the  draftsman  has  at  his  command 


of  the  45-degree  triangle,  as  A  E,  is  placed  against  the 
blade  of  the  T-square,  and  the  vertical  division  of  the 
circle  is  drawn  along  the  other  short  side  C  E. 

In  Fig.  178  the  vertical  and  horizontal  divisions 
of  the  circle,  A  B  and  C  D,  are  drawn  as  before,  after 
which  one  of  the  shorter  sides  of  the  45 -degree  triangle 
is  placed  against  the  T-square,  and  the  long  or  oblique 
side  E  F  is  brought  to  the  center  of  the  circle  and 
another  division  G  I  is  drawn.  By  reversing  the  posi- 
tion of  the  triangle  the  last  division  II  K  is  drawn, 
thus  dividing  the  circle  into  eight  equal  parts. 

In  Fig.  179,  after  drawing  the  divisions  A  B  and 
C  D  as  before,  the  30  X  60-degree  triangle  is  placed  in 
the  position  shown  at  A  E  F,  and  the  division  E  N  is 
drawn  along  its  hypothenuse  or  oblique  side.  Bv  re- 
versing the  position  of  the  triangle,  still  keeping  the 


Fig.  177.— Circle  Divided  into  Four 
Equal  Parts  by  the  Use  of  a  Triangle 
or  Set-Square. 


Fig.  178.— Circle  Divided  into  Eight 
Equal  Parts  by  the  Use  of  a  45-degree 
Triangle. 


Fig.  170.— Circle  Divided  into  Twelve, 
Equal  Parts  by  the  Use  of  a  80  x  60- 
degree  Triangle. 


the  means  of  drawing  lines  at  angles  of  90,  60,  45  and 
30  degrees,  and  by  combination  75  degrees  (45  -)-  30) 
and  15  degrees  (90  —  75).  With  the  45-degree 
angle  he  can  bisect  a  right  angle,  and  with  the  30  and 
60-degree  angles  he  can  trisect  it. 

The  pattern  draftsman  sometimes  finds  it  con- 
venient to  have  a  set-square  in  which  the  sharpest 
angle  is  one  of  22£  degrees  (one-half  of  45)  for  use  in 
drawing  the  octagon  in  a  certain  position  which  will 
be  referred  to  later. 

In  Figs.  177,  178,  179  and  180  are  illustrated 
the  application  of  the  foregoing,  in  which  the  circle  is 
divided,  by  the  use  of  the  triangles  above  described, 
into  four,  eight  and  twelve  equal  parts.  In  Fig.  177 
the  horizontal  division  A  B  of  the  circle  is  drawn  by 
means  of  the  T-square  placed  against  the  side  of  tin- 
drawing  board,  after  which  one  of  the  shorter  sides 


side  A  F  against  the  blade  of  the  T-square,  the  divi- 
sion J  K  may  be  drawn.  Changing  the  position  of  the 
triangle  now  so  that  its  shortest  side  comes  against  the 
blade  of  the  T-square,  as  shown  dotted  at  G  H  F,  the 
division  G  M  is  drawn,  and  again  reversing  its  posi- 
tion, still  keeping  its  shortest  side  against  the  T-square, 
the  last  division  I  L  may  be  drawn,  thus  dividing  the 
circle  into  twelve  equal  parts. 

In  Fig.  180  the  circle  is  divided  into  eight  equal 
parts,  but  differing  from  that  shown  above  in  this 
respect  that,  while  in  Fig.  178  two  of  the  divisions  lie 
parallel  with  the  sides  of  the  drawing  board,  in  the 
latter  case  none  of  the  divisions  are  parallel  with  t lie- 
sides  of  the  board  or  can  be  drawn  with  the  f-square; 
but  if  this  method  is  used  in  drawing  an  octagon,  as 
shown  dotted  in  Fig.  180,  four  of  the  sides  of  the  oc- 
tagon can  be  drawn  with  the  T-square  illRl  tue  other 


Geometrical 


51 


four  with  the  45-degree  triangle.  The  position  of  the 
-•2$  X  67-i-degree  triangle  in  drawing  tlie  divisions  of 
the  circle  is  shown  at  A  15  (..'  and  I.)E  C,  while  the  posi- 
tion of  the  45-degree  triangle  in  drawing  the  oblique 
sides  of  an  octagon  figure  is  shown  at  F.  It  will  thus 
be  seen  that  the  22£  X  tiT^-degree  triangle  is  available 
in  drawing  accurately  the  miter  line  for  all  octagon 
miters. 

As  a  triangle  in  whatever  form  it  may  be  con- 
structed is  intended  to  be  used  by  sliding  it  against, 
the  blade  of  the  T-^quare.  "11  the  angles  above  men- 
tioned are  calculated  with,  reference  to  the  lines  drawn 
by  the  T-square.  In  practical  use  it  will  be  found  in- 
convenient in  drawing  such  lines  to  actually  bring  the 
point  of  a  set-square  to  the  center  of  a  circle.  A 
better  method,  and  one  which  makes  use  of  the  same 
principles,  is  shown  in  Fig.  181.  The  blade  of  the 


Fig.    ISO.— Circle   Divided    into   Eight   Equal    Parts  by   the 
Use  of  a  1S&%  x  Vt^-degree  Triangle, 

T-square  is  placed  tangent  to  or  near  the  circle,  as  shown 
by  A  B.  One  side  of  a  45-degree  triangle  is  placed 
against  it,  as  shown,  its  side  C  F  being  brought  against 
the  center.  The  line  C  F  is  then  drawn.  By  reversing 
the  trrangle,  as  shown  by  the  dotted  lines,  the  line  E 
D  is  drawn  at  right  angles  to  C  F,  thus  dividing  the 
circle  into  quarters. 

A  similar  use  of  the  30  X  60-degree  triangle  is 
shown  in  Fig.  182,  by  which  a  circle  is  divided  into 
six  equal  parts.  Bring  the  blade  of  the  T-square 
tangent  to  or  near  the  circle,  as  shown  by  A  B.  Then 
place  the  set-square  as  shown  by  G  B  M,  bringing  the 
side  G  B  against  the  center  of  the  circle,  drawing  the 
line  D  L.  Then  place  it  as  shown  by  the  dotted  lines, 
bringing  the  side  A  H  against  the  center,  scribing  the 
line  F  E.  Then,  by  reversing  the  set-square,  placing 
the  side  G  M  against  'the  straight-edge,  erect  the  per- 


pendicular C  I,  completing  the  division.  The  follow- 
ing are  a  few  of  the  problems  to  which  these  principles 
may  be  advantageously  applied. 

48.  To  Inscribe  an  Equilateral  Triangle  within  a  Given 
Circle. — In  Fig.  183,  let  D  be  the  center  of  the  given 
circle.  Set  the  side  E  F  of  a  30-degree  set-square 
against  the  T-square,  as  shown,  and  move  it  along 


D  F 

Fig.  181.— Proper  Method  of  Using  a  45-degree  Triangle. 

until  the  side  E  G  touches  D.  Mark  the  point  B  upon 
the  circumference  of  the  circle.  Eeverse  the  set-square 
so  that  the  point  E  will  come  to  the  right  of  the  side 
F  G  and  move  it  along  in  the  reversed  position  until 
the  side  E  G  again  meets  the  point  D,  and  mark  the 
point  C.  Now  move  the  T-square  upward  until  it 
touches  the  point  D,  and  mark  the  point  A.  Then 
A  B  and  C  are  points  which  divide  the  circle  into  three 
equal  parts.  The  triangle  may  be  easily  completed 
from  this  stage  by  drawing  lines  connecting  A  B,  B  C 
and  C  A,  with  any  straight-edge  or  rule,  but  greater 
accuracy  is  obtained  by  the  further  use  of  the  set- 
si  [uare,  as  follows :  Place  the  side  F  G  of  the  set- 
square  against  the  T-square,  as  shown  in  Fig.  184, 


Fig.  182  —Method  of  Using  a  30  x  60-degree  Triangle  in 
Dividing  the  Circle. 


and  move  it  along  until  the  side  E  G  touches  the  points 
A  and  C,  as  shown.  Draw  A  C,  which  will  be  one 
side  of  the  required  triangle.  Set  the  side  E  F  of  the 
set-square  against  the  T-square,  and  move  it  along  until 
the  side  F  G  coincides  with  the  points  C  and  B.  Then 
draw  C  B,  which  will  be  the  second  side  of  the  triangle. 


52 


The  New  Metal   Worker  Pattern  Book. 


Place  the  side  F  G  of  the  set-square  against  the  T- 
square,  with  the  side  E  F  to  the  right,  and  move  it 
along  until  the  side  E  G  coincides  with  the  points  A 
and  B.  Then  draw  A  B,  thus  completing  the  figure. 
The  same  results  may  be  accomplished  with  less  work 
by  first  establishing  the  point  A  by  bringing  the 
T-square  against  the  center,  and  then  using  the  set- 
square,  as  shown  in  Fig.  184.  The  different  methods 
are  here  given  in  order  to  more  clearly  illustrate  the 
use  of  the  tools  employed. 


Fig.  183. 


Fig.  184. 

To  Inscribe  an  Equilateral  Triangle  within  a 
Given  Circle. 


49.  To  Inscribe  a  Square  within  a  Given  Circle  - 

Let  D,  in  Fig.  185,  be  the  center  of  the  given  circle. 
Place  the  side  E  F  of  a  45-degree  set-square  against 
the  T-square,  as  shown,  and  move  it  along  until  the 
side  E  G  meets  the  point  D.  Mark  the  points  A  and 
B.  Eeverse  the  set-square,  and  in  a  similar  manner 
mark  the  points  C  and  H.  The  points  A,  H,  B  and 
C  are  corners  of  the  required  square.  Move  the  J- 
square  upward  until  it  coincides  with  the  points  A  and 
H  and  draw  A  H,  as  shown  in  Fig.  186.  In  like  man- 
ner draw  C  B.  With  the  side  E  F  of  the  set-square 
against  the  T-square,  move  it  along  until  the  side  G  F 


coincides  with  the  points  B  and  H,  and  draw  B  H.     In 

a  similar  manner  draw  C  A,  thus  completing  the  figure. 

50.  To  Inscribe  a  Hexagon  within  a  Given  Circle.— 

In  Fig.  187,  let  0  be  the  center  of  the  given  circle. 
Place  the  side  E  F  of  a  30-degree  set-square  against 
the  T-square,  as  shown.  Move  the  set-square  along 
until  the  side  E  G  meets  the  point  0.  Mark  the  points 
A  and  B.  Reverse  the  set-squaiv,  and  in  like  manner 
mark  the  points  C  and  D.  With  the  side  F  G  of  the 
set-square  against  the  1  -square,  move  it  along  until 


Fig.  185. 


Fig.  186. 
To  Inscribe  a  Square  within  a  Given  Circle. 

the  side  E  F  meets  the  point  0,  and  mark  I  and  H. 
Then  A,  H,  D,  B,  I  and  C  represent  the  angles  of  the 
proposed  hexagon.  From  this  stage  the  figure  may  be 
readily  finished  by  drawing  the  sides  by  means  of  these 
points,  using  a  simple  straight-edge;  but  greater  ac- 
curacy is  attained  in  completing  the  figure  by  the 
further  use  of  the  set-square,  as  shown  in  Fig.  188. 
With  the  side  E  F  of  the  set-square  against  the 
f -square,  as  shown,  draw  the  line  II  D,  and  by  mov- 
ing the  T-square  upward  draw  the  side  C  I.  Reversing 
the  set-square  so  that  the  point  F  is  to  the  left  of  the 
point  E,  draw  the  side  A  H,  and  also,  by  shifting  the 


Geometrical    Problems. 


53 


T-square,  the  side  I  B.  With  the  edge  E  F  of  the 
set-square  against  the  T-square,  move  it  up  until  the 
side  G  F  coincides  with  the  points  B  and  D,  and  draw 
the  side  B  D.  In  like  manner  draw  A  C,  thus  com- 
pleting the  figure.  In  this  figure,  as  with  the  triangle, 
the  same  results  may  be  reached  by  establishing  the 
points  H  and  I,  by  means  of  a  diameter  drawn  at  right 
angles  to  the  T-square,  as  shown  in  the  engravings, 
and,  using  it  as  a  base,  employing  the  set-square,  as 
shown  in  Fig.  188.  The  first  method  shown  is,  how- 


Fig.  18?. 


Fig.  188. 
To  Inscribe  a  Regular  Hexagon  within  a  Given  Circle. 

ever,  to  be  preferred  in  many  instances,  on  account  of 
its  great  accuracy. 

51.  To  Inscribe  an  Octagon  within  a  Given  Circle.— 

In  Fig.  189,  let  K  be  the  center  of  the  given  circle. 
Place  a  45-degree  set-square  as  shown  in  the  engraving, 
bringing  its  long  side  in  contact  with  the  center,  and 
mark  the  points  E  and  A.  Keeping  it  in  the  same  po- 
sition, move  it  along  until  its  vertical  side  is  in  contact 
with  K  and  mark  the  points  D  and  II.  Reverse  the 
set-square  from  the  position  shown  in  the  engraving, 
and  mark  the  points  C  and  G.  Move  the  T-square  up- 
ward until  it  touches  the  point  K,  and  mark  the  points 


B  and  F.  Then  A,  H,  G,  F,  E,  D,  C  and  B  are  cor- 
ners of  the  octagon.  The  figure  may  now  be  readily 
completed  by  drawing  the  sides,  by  means  of  these 
points,  using  any  rule  or  straight-edge  for  the  purpose, 


Fig.  189.— To  Inscribe  a  Regular  Octagon  within  a 
Given  Circle. 

all  as  shown  by  A  H,  H  G,  G  F,  F  E,  E  D,  D  C,  C  B 
and  B  A,  or  by  means  of  a  22£  X  67^-degree  set- 
square. 

52.  To  Draw  an  Equilateral  Triangle  upon  a  Given 

Side.— In  Fig.  190,  let  A  B  be  the  given  side.  First 
bring  the  line  A  B  at  right  angles  to  the  blade  of  the 
T-square.  Then  set  the  edge  C  B  of  a  30-degree  set- 
square  against  the  T-square,  and  move  it  along  until 
the  edge  B  D  meets  the  point  B,  and  draw  the  line 
B  F.  Reverse  the  set-square,  still  keeping  the  side 
C  B  against  the  T-square,  and  move  it  along  until  the 


Fig.  190.— To  Draw  an  Equilateral  Triangle  upon  a 
Given    Side. 


side  B  D  meets  the  point  A,  and  draw  the  line  A  F, 
thus  completing  the  figure. 

53.  To  Draw  a  Square  upon  a  Given  Side.— In  Fig. 
191,  let  A  B  be  the  given  side  drawn  at  right  angles 


54 


Tin    \':/C    M'lttl     II »/•/•''/•    Pntt'-ni    ttouk. 


to  the  blade  of  the  T-square.  Set  the  edge  E  F  of  a 
45-degree  set-square  against  the  T-square,  as  shown, 
and  move  it  along  until  the  side  E  G  meets  the  point 
B,  and  draw  B  I  indefinitely.  Reverse  the  set-square, 
and,  bringing  the  side  E  G  against  the  point  A,  draw  A 


Fig.  191. — To  Draw  a  Square  -upon  a  Given  Side. 

F  indefinitely.  Bring  the  T-square  against  the  point 
B  and  draw  B  F,  producing  it  until  ,it  meets  the  line 
A  F  in  the  point  F.  In  like  manner  draw  A  I,  meet- 
ing the  line  B  I  in  the  point  I.  Then,  with  the  set- 
square  placed  as  shown  in  the  engraving,  connect  I 
and  F,  thus  completing  the  required  figure. 

54.  To  Draw  a  Regular  Hexagon  upon  a  Given  Side. 
— In  Fig.  192,  let  A  B  be  the  given  side  in  a  vertical 
position.  Set  the  edge  G  H  of  a  30-degree  set-square 
against  the  T-square,  as  shown,  and  move  it  along 


Fig.  192. — To  Draw  a  Regular  Hexagon  upon  a  Given  Side. 

until  the  edge  I  G  coincides  with  the  point  A,  and 
draw  the  line  A  D  indefinitely.  Reverse  the  set- 
square,  still  keeping  the  edge  G  H  against  the 
T-square,  and  move  it  along  until  the  side  I  G  coin- 
cides with  the  point  B,  and  draw  B  E  indefinitely. 
These  lines  will  intersect  in  the  point  0,  which  will  be 


the  center  of  the  required  figure.  Still  keeping  the 
edge  G  H  of  the  set-square  against  the  T-square,  move 
it  along  until  the  perpendicular  edge  I  II  meets  the 
point  O,  and  through  O  draw  F  C  indefinitely.  With 
the  set-square  in  the  position  shown  in  the  engraving 
slide  it  along  until  the  edge  I  G  meets  the  point 
B,  and  draw  B  C,  producing  it  until  it  meets  the  line 
F  C  in  the  point  C.  Reverse  the  set-square,  still 
keeping  the  edge  G  H  against  the  T-square,  and  draw 
the  line  C  D,  producing  it  until  it  meets  the  line  A  D 
in  the  point  D.  Slide  the  set-square  along  until  the 
side  I  H  meets  the  point  D,  and  draw  the  line  D  E, 
meeting  the  line  B  E  in  the  point  E.  Move  the  set- 
square  along  until  the  edge  I  G  meets  the  point  A,  and 
draw  the  line  A  F,  meeting  the  line  C  F  in  the  point 
F.  Now  bring  the  set-square  to  its  first  position  and 


Fig.  193.— To  Draw  a  Regular  Octagon  upon  a  Given  Side. 

slide  it  along  until  the  edge  I  G  meets  the  points  F  and 

E,  and  draw  F  E,  thus  completing  the  required  figure. 

55.  To  Draw  a  Regular  Octagon  upon  a  Given  Side.— 

In  Fig.  193,  let  C  D  be  the  given  side,  drawn  perpen- 
dicular to  the  blade  of  the  T-square.  Place  one  of  the 
short  sides  of  a  45-degree  set-square  against  the 
T-square,  as  shown  in  the  engraving.  Move  the  set- 
square  along  until  its  long  side  coincides  with  the 
point  C.  Draw  the  line  C  B,  and  make  it  in  length 
equal  to  C  D.  With  the  T-square  draw  the  line  A  B, 
also  in  length  equal  to  C  D.  Reverse  the  set-square, 
and  bring  the  edge  against  the  point  A.  Draw  A  II 
in  length  the  same  as  C  D.  Still  keeping  a  short  side 
of  the  set-square  against  the  T-square,  slide  it  along 
until  the  other  short  side  meets  the  point  H,  and  draw 
H  G,  also  of  the  same  length.  Then,  using  the  long 
side  of  the  set-square,  draw  G  F  of  corresponding 
length.  By  means  of  the  T-square  draw  F  E,  and  by 
reversing  the  set-square  draw  E  D,  both  in  length 


Geometrical 


equal  to  the  original  side,  C  D,  joining  it  in  tin;  point 
D,  thus  completing  the  required  octagon. 

56.  To  Draw  an  Equilateral  Triangle  about  a  Given 
Circle.— In  Fig.  194,  let  ()  lie  the  center  of  the  given 
circle.  Place  the  edge  E  F  of  a  30-dcgree  set-square 
against  the  T-square,  !IS  shown,  and  move  it  along  un- 
til the  edge  F  G  meets  the  center  O,  and  mark  the 
point  A  upon  the  circumference  of  the  circle.  Reverse 
the  set-square,  still  keeping  the  edge  E  F  against  the 
T-sqnare,  and  in  like  manner  mark  tin-  point  B.  Move 


Fig.  191. 


Fig.  195. 
To  Draw  an  E<iuilateral  Triangle  about  a  Given  Circle. 

the  T-square  upward  until  it  meets  the  point  O,  and 
mark  the  point  C.  The  required  figure  will  be  de- 
scribed by  drawing  lines  tangent  to  the  circle  at  the 
points  A,  B  and  C,  which  may  be  done  in  the  manner 
following,  as  indicated  in  Fig.  195  Place  the  edge 
E  G  of  the  set-square  against  the  T-square,  and  slide  it 
along  until  the  edge  F  G  touches  the  circle  in  the  point 
B.  Draw  I  K  indefinitely.  Reverse  the  set-square, 
keeping  the  same  edge  against  the  T-S(luare,  and  move 
it  along  until  its  edge  F  G  touches  the  circle  in  the 
point  A,  and  draw  I  L,  intersecting  I  K  in  the  point  I, 


the  other  end  being  indefinite.  Then,  placing  the  edge 
F  E  of  the  set-square  against  the  T-square,  bring  its 
edge  E  G  against  the  circle  in  the  point  C,  and  draw 
L  K,  intersecting  I  D  in  the  point  L  and  I  K  in  the 
point  K,  thus  completing  the  figure.  The  first  part  of 
tins  operation  is  not  really  necessary.  The  sides  of  the 
set-square  simply  can  be  brought  tangent  to  the  circle, 
as  in  Fig.  195. 

57.  To  Draw  a  Hexagon  about  a  Given  Circle.— In 
Fig.  1 96,  let  0  be  the  center  of  the  given  circle.     Place 


Fig.  196. 


Fig.  197. 
To  Draw  a  Hexagon  about  a  Givtn  Circle. 

the  edge  E  F  of  a  30-degree  set-square  against  the 
J-square,  and  slide  it  along  until  the  edge  F  G  meets  the 
point  0,  and  mark  the  points  B  and  A.  Reverse  the  set- 
square,  still  keeping  the  edge  E  F  against  the  T-square, 
and  in  like  manner  mark  the  points  C  and  D.  Bring 
the  edge  of  the  T-square  against  O,  and  mark  the  points 
1  and  K.  Then  C,  A,  K,  D,  B  and  I  are  six  points  in 
the  circumference  of  the  circle,  corresponding  to  the 
six  sides  of  the  required  figure.  The  hexagon  is  com- 
pleted by  drawing  a  side  tangent  to  the  circle  at  each 
of  these  several  points,  which  may  be  done  by  using 


56 


Tin'   \*  //' 


Worker  Pattern  Book. 


the  set-square  as  follows,  and  as  shown  in  Fig.  197. 
With  the  edge  E  G  of  the  set-square  against  the 
T-square,  bring  the  edge  F  G  against  the  circle  at  the 
point  C,  as  shown,  and  draw  L  M  indefinitely.  Re- 
verse the  set-square,  and  in  like  manner  bring  it  against 
the  circle  at  the  point  A,  and  draw  M  N,  cutting  L  M 
in  the  point  M,  and  extending  indefinitely  in  the  direc- 
tion of  N.  Slide  the  set-square  along  until  the  edge 
E  F  meets  the  circle  in  the  point  K,  and  draw  N  P, 
intersecting  M  N  in  the  point  N,  and  extending  in  the 


Fig.  198. 


Fig.  199. 
To  Draw  an  Octagon  about  a  Given  Circle. 

direction  of  P  indefinitely.  With  the  set-square  in 
its  first  position  slide  it  along  until  the  edge  F  G 
meets  the  circle  in  the  point  D,  and  draw  R  P,  cut- 
ting N  P  in  the  point  P,  but  being  indefinite  in 
the  direction  of  R.  Reverse  the  set-square,  and  in 
like  manner  draw  R  S  tangent  to  the  circle  in  the 
point  B,  cutting  P  R  in  the  point  R,  and  extending  in 
the  direction  of  S  indefinitely.  Slide  the  set-square 
along  until  its  edge  E  F  meets  the  circle  in  the  point 
I,  and  draw  S  L,  cutting  R  S  in  the  point  S  and  L  M 
in  the  point  L,  thus  completing  the  required  figure. 


In  this  problem,  as  in  the  previous  one,  if  care  be  taken 
the  first  part  of  the  operation  can  be  dispensed  with  bv 
simply  placing  the  triangle  in  proper  position  and  draw- 
ing the  sides  of  the  figure  tangent  to  the  circle,  as 
shown  in  Fig.  197. 

58.  To  Draw  an  Octagon  about  a  Given  Circle.— In 
Fig.  198,  let  O  be  the  center  of  the  given  circle.  With 
the  edge  E  F  of  a  45-degree  set-square  against  the  T- 
square,  as  shown,  move  it  along  until  the  side  E  G 
meets  the  point  O,  and  mark  the  points  A  and  B. 
Reverse  the  set-square,  and  in  like  manner  mark  the 
points  C  and  D.  Slide  the  set-square  along  until  the 
vertical  side  G  F  meets  the  point  0,  and  mark  the 
points  H  and  I.  Move  the  T-square  up  until  it  meets 
the  point  0,  and  mark  the  points  K  and  L.  Then  A, 
I,  D,  L,  B,  H,  C  and  K  are  points  in  the  circumfer- 
ence of  the  given  circle  corresponding  to  the  sides  of 


Fig.  £00.— To  Draw  a  Square  about  a  Given  Circle. 

the  required  figure.  The  octagon  is  then  to  be  com- 
pleted by  drawing  lines  tangent  to  the  circle  at  these 
several  points,  as  shown  in  Fig.  199,  which  may  be 
done  by  the  use  of  the  set-square,  as  follows :  With 
the  edge  E  F  of  the  set-square  against  the  T-square, 
as  shown,  bring  the  edge  E  G  against  the  circle  in  the 
point  D,  and  draw  M  N  indefinitely.  Sliding  the  set- 
square  along  until  the  vertical  edge  F  G  meets  the 
circle  in  the  point  L,  draw  N  P,  cutting  M  N  in  the 
point  N,  and  extending  in  the  opposite  direction  in- 
definitely. Reverse  the  set-square,  and  bringing  the 
edge  E  G  against  the  circle  in  the  point  B,  draw  P  R, 
cutting  N  P  in  the  point  P,  and  extending  indefinitely 
in  the  direction  of  R.  Move  the  T-square  upward  un- 
til it  meets  the  circle  in  the  point  H,  and  draw  the  line 
S  R,  meeting  P  R  in  the  point  R,  and  extending  in- 
definitely in  the  opposite  direction.  Then,  with  the 
set-square  placed  as  shown  in  the  engraving,  move  it 


Problems. 


until  its  edge  E  (f  meets  the  circle  in  the  point  C,  and 
draw  S  T,  meeting  S  R  in  the  point  S,  and  continuing 
indefinitely  in  the  direction  of  T.  With  the  set-square 
in  the  same  position,  move  it  along  until  its  edge  G  F 
meets  the  circle  in  the  point  K,  and  draw  T  U,  cutting 
S  T  in  the  point  T,  and  extending  in  the  opposite  di- 
rection indefinitely.  Reverse  the  set-square,  and  bring- 
ing its  long  side  against  the  circle  in  the  point  A,  draw 
U  V,  cutting  T  U  in  the  point  U,  and  continuing  in- 
definitely in  the  opposite  direction.  Bring  the  T-- 
square against  the  circle  in  the  point  I,  and  draw  V  M, 
connecting  U  V  and  M  N  in  the  points  V  and  M  re- 
spectively, thus  completing  the  figure.  The  above  rule 
will  be  found  very  convenient  for  use,  although,  as 
the  student  may  discover,  the  first  part  of  the  opera- 
tion is  not  absolutely  necessary. 

59.  To  Draw  a  Square  about  a  Given  Circle.— In 
Fig.  200,  let  0  be  the  center  of  the  given  circle.    Place 


Fig.  SOL— To  Draw  a  Square  upon  a  Given  Side. 

the  blade  of  the  T-square  against  the  point  0,  and 
draw  the  line  A  O  B.  With  one  of  the  shorter  sides 
E  F,  of  a  45-degree  set-square  against  the  T-sc[uare? 
and  with  the  other  short  side  against  the  point  0,  draw 
the  line  DOC.  Move  the  T-square  upward  until  it 
strikes  the  point  C,  and  draw  the  line  II  C  I.  Move 
it  down  until  it  strikes  the  point  D,  and  draw  the  line 
E  D  K.  With  the  side  E  F  of  the  set-square  against 
the  T-square,  as  shown  in  the  engraving,  bring  the  side 
E  G  against  the  point  A,  and  draw  E  A  II.  In  like 
manner  bring  it  against  the  point  B,  and  draw  K  B  I, 
thus  completing  the  figure.  It  is  to  be  observed  that 
the  several  lines  composing  the  sides  of  the  square  are 
tangent  to  the  circle  in  the  points  A  C  B  and  D  re- 
spectively. The  only  object  served  by  drawing  the 
diameters  A  B  and  C  D  is  that  of  obtaining  greater 
accuracy  in  locating  the  points  of  tangency. 

60.  To  Draw  a  Square  upon  a  Given  Side.— Let  A  B 
o.f  Fig.  201  be  the  given  side  placed  parallel  to  one 


side  of  the  drawing  board.     Place  one  of  the  shorter 
edges  of  a  45-degree  set-square  against  the  T-square,  as 
placed  for  drawing  the  given  side,  and  slide  it  along 
until  the  long  edge  touches  the  point  A,  and  draw  the 
diagonal  line  A  C  indefinitely.      Place  the  T-square  so 
that  its  head  comes  against  the  left  side  of  the  board, 
as  shown  by  the  dotted  lines  in  the  engraving,  and, 
bringing  the  blade  against  the  point  A,  draw  A  D  in- 
definitely.     Then  bringing  the  blade  against  the  point 
B,  draw  B  C,  stopping  this  line  at  the  point  of  inter- 
section with  the  line  A  C,  as  shown  at  C.      Bring  the 
T-square  back  to  the  original  position  and  draw  the 
line  C  D,  thus  completing  the  figure.     In  the  case  of 
a  large  drawing  board,  unless  the  figure  is  to  be  located 
very  near  one  corner  of  it,  or  in  the  case  of  a  drawing 
board  of  which  the    adjacent    sides    are  not  at   right 
angles,  it  will  be  desirable  to  use  the  right  angle  of  the 
set-square,  instead  of  changing  the  T-square  from  one 
side  to  the  other,  as  above  described.     The  object  of 
drawing  the  diagonal  line  A  C  is    to  determine    the 
length  of  the  side  C  B.     This  also  may  be  done  by  the 
use  of  the  compasses  instead  of  the  set-square,  as  fol- 
lows :    From  B  as  center,  with  B  A  as  radius,  describe 
the  arc  A  O  C.      Place  the  T-square  as  shown   by  the 
dotted  lines,  and,  bringing  it  against  the  point  B,  draw 
B  C,  producing  it  until  it  intercepts  the  arc  A  O  C  in 
the  point  C.     The  remaining  steps  are  then  to  be  taken 
in  the  manner  above  described. 

ffl.— BY   MEANS  OF   THE   PROTRACTOR. 

The  protractor,  which  has  been  already  described 
and  illustrated  (see  Fig.  116,  Chapter  II),  is  an  instru- 
ment for  measuring  angles.  The  usual  form  of  this 
instrument  is  a  semicircle  with  a  graduated  edge,  the 
divisions  being  more  or  less  numerous,  according  to  its 
size.  In  instruments  of  ordinary  size  the  divisions  are 
single  degrees,  numbered  by  5s  or  by  10s,  while  in 
larger  sizes  the  divisions  are  made  to  fractions  of 
degrees. 

Since  the  protractor  by  its  construction  affords  the 
means  of  measuring  or  of  setting  off  any  angle  whatso- 
ever, it  is  especially  useful  in  circumscribing  or  in- 
scribing polygons,  or  of  erecting  them  upon  a  given 
side.  As  its  use  is  of  infrequent  occurrence  among 
pattern  draftsmen,  only  a  few  problems  in  inscribing 
will  be  given,  which  will  be  sufficient  to  enable  the 
reader  to  apply  it  in  other  cases  that  may  arise. 

61.  To  Inscribe  an  Equilateral  Triangle  within  a  Given 

;  Circle.— In  Fig.  202,  let  0  be  the  center  of  the  given 

circle.      Through  0  draw  a  diameter,   as  shown  by 


58 


T/ie  New  Metal    ll'w/r/-   I '„(/<;•„ 


COD.  Place  the  protractor  so  that  its  center  point 
shall  coincide  with  O,  and  turn  it  until  the  point  mark- 
ing 60  degrees  falls  upon  the  line  C  0  I).  Then  mark 
points  in  the  circumference  of  the  circle  corresponding 
to  0  (zero)  and  120  degrees  of  the  protractor,  as  shown 
by  B  and  E  respectively.  Draw  the  lines  C  E,  E  B 
and  B  C,  thus  completing  the  required  figure.  The 
reasons  for  these  several  steps  are  quite  evident.  The 
circle  consists  of  360  degrees.  Then  each  side  of  an 
equilateral  triangle  must  represent  one-third  of  360 
degrees,  or  120  degrees.  Assume  the  point  C  for  one 
of  the  angles,  and  draw  the  line  COD.  Then,  by  the 
nature  of  the  figure  to  be  drawn,  D  must,  fall  opposite 
the  center  of  one  side.  Therefore,  since  60  is  the  half 
of  120  (the  length  of  one  side  in  degrees),  place  60 
opposite  the  point  D,  and  mark  0  and  120  for  the  other 
angles,  then  complete  the  figure  by  drawing  the  lines 


Fig.  202.— To  Inscribe  an  Equilateral  Triangle  within  a 
Given  Circle. 

as  shown.  Since  in  many  cases  the  protractor  is  much 
smaller  than  the  circle  in  which  the  figure  is  to  be  con- 
structed, it  becomes  necessary  to  mark  the  points  at 
the  edge  of  the  instrument,  and  carry  them  to  the  cir- 
cumference by  drawing  lines  from  the  center  of  the 
circle  through  the  points,  producing  them  until  the 
circle  is  reached. 

62.  To  Inscribe  a  Square  within  a  Given  Circle.— In 
Fig.  203,  let  0  be  the  center  of  the  given  circle. 
Through  0  draw  a  diameter,  as  shown  by  COD. 
Place  the  protractor  so  that  its  center  point  coincides 
with  0,  and  turn  it  until  the  point  marking  45  degrees 
falls  upon  the  line  COD.  Mark  points  in  the  circum- 
ference of  the  circle  corresponding  to  0,  90  and  180 
degrees  of  the  protractor,  as  shown  by  F,  G  and  E  re- 
spectively. From  G,  through  the  center  0,  draw  G  O 
H,  cutting  the  circumference  of  the  circle  in  the  point 


II.  Then  E,  G,  F  and  II  are  the  angles  of  the  required 
figure,  which  is  to  be  completed  by  drawing  the  sides 
E  G,  G  F,  F  H  and  II  E.  Since  the  circle  is  composed 
of  360  degrees,  one  side  of  an  inscribed  square  must 
represent  one-fourth  part  of  360  degrees,  or  90  degrees. 
The  half  of  90  degrees  is  45  degrees.  Hence,  in  set- 
ting the  protractor,  the  point  representing  45  degrees 
was  placed  opposite  the  point  in  which  it  is  desired  the 
center  of  one  of  the  .sides  shall  fall,  or,  in  other  words, 
upon  the  line  COD.  Then,  having  marked  points  90 
degrees  removed  from  each  other,  or,  as  explained 
above,  opposite  the  points  0,  90  and  180  of  the  pro- 
tractor, as  shown  by  F,  G  and  E,  the  fourth  point  was 
obtained  by  the  diagonal  line.  It  is  evident  that  II 
must  fall  opposite  G,  upon  a  line  drawn  through  the 
center.  Or  the  protractor  might  have  been  mo  veil 
around,  and  a  space  of  90  degrees  measured  from  either 


Fig.  SOS. — To  Inscribe  a  Square  within  a  Oiven  Circle. 

F   or  E,  which,  as  will  be  clearly  seen,  would  have 
given  the  same  point,  H. 

63.  To  Inscribe  an  Octagon  within  a  Given  Circle.— 

Through  the  center  0  of  the  given  circle,  Fig.  204, 
draw  a  diameter,  A  0  B,  upon  which  the  center  of 
one  side  is  required  to  fall.  Place  the  protractor  so 
that  its  center  point  shall  coincide  with  the  center  O, 
and  turn  it  so  that  the  point  representing  22£  degrees 
shall  fall  on  the  line  A  O  B.  Then  mark  points  in 
the  circumference  of  the  circle  corresponding  to  0,  45, 
90,  135  and  180  degrees  of  the  protractor,  as  shown 
by  E,  G,  H,  I  and  F.  Reverse  the  protractor,  and  in 
like  manner  mark  the  points  M,  Land  K;  or  these 
points  may  be  obtained  by  drawing  lines  from  I,  II 
and  G  respectively  through  the  center  O,  cutting  the 
circumference  in  M,  L  and  K.  The  figure  is  to  be 
completed  by  drawing  the  sides  F  I,  I  H,  II  G,  G  E, 


Geometrical    Problems. 


no 


E  M,  M  L,  L  K  and  K  F.  Since  the  circle  consists 
of  360  degree's,  one  side  of  an  octagon  must  represent 
45  degrees,  or  oiie-eiglit.li  of  360.  The  half  of  45  is 
•!•!-}>.  Hence,  the  point  of  the  protractor  representing 
22£  degrees  was  placed  upon  the  line  A  0  B,  which 
represents  the  center  of  one  side  of  the  required  figure. 
Having  thus  established  the  position  of  one  side,  the 
other  sides  of  the  figure  are  located  by  marking  points 
in  the  circumference  of  the  circle  opposite  points  in  the 
protractor  at  regular  intervals  of  45  degrees. 

64.  To  Inscribe  a  Dodecagon  within  a  Given  Circle.— 
In  Fig.  205,  let  O  be  the  center  of  the  given  circle. 


Fig.  £04.— To  Inscribe  an  Octagon  within  a  Given  Circle. 

Through  0  draw  the  diameter  A  O  B,  at  right  angles 
to  which  one  of  the  sides  of  the  polygon  is  required  to 
be.  Set  the  protractor  so  that  the  center  point  of  it 
coincides  with  the  center  O,  and  revolve  it  until  the 
point  marking  15  degrees  falls  upon  the  line  A  0  B. 
With  the  protractor  in  this  position,  mark  points  in 
the  circumference  of  the  circle  opposite  the  points  in 
the  protractor  representing  0,  30,  60,  90,  120,  150  and 
180  degrees,  as  shown  by  E,  F,  G,  H,  I,  K  and  L. 
Then  these  points  will  represent  angles  of  the  required 
polygon.  The  remaining  angles  may  be  obtained  by 
placing  the  protractor  in  like  position  in  the  opposite 
half  of  the  semicircle,  or  they  may  be  determined  by 
drawing  lines  from  the  points  F,.  G,  H,  I  and  K 
through  the  center  O,  producing  them  until  they  cut 


the  circumference  in  the  points  M,  N,  P,  R  and  S, 
which  are  the  remaining  angles.  The  figure  is  now  to 
be  completed  by  drawing  the  sides,  as  shown.  In  a  do- 
decagon, or  twelve-sided  figure,  each  side  must  occupy 
a  space  represented  by  one-twelfth  of  360  degrees,  or 
30  degrees  of  the  protractor.  As  the  side  F  E  was  re- 
quired to  be  located  in  equal  parts  upon  opposite  sides 
of  A  0  B,  the  middle  of  one  division  of  the  protractor 
representing  a  side  (that  is,  15  degrees,  or  one-half  of 
30  degrees)  was  placed  upon  the  line  A  O  B.  Having 
thus  established  the  position  of  one  side,  the  others 
are  measured  off  in  manner  above  described. 


p  R 

Fig.  t05. — To  Inscribe  a  Dodecagon  within  a  Given  Circle. 

In  making  use  of  the  protractor  to  erect  a  regular 
polygon  upon  a  given  side,  the  exterior  angle,  or  angle 
formed  by  an  adjacent  side  with  the  given  side  ex- 
tended, as  E  B  D  in  Figs.  168,  170  and  171,  is  found 
by  dividing  360  degrees  by  the  number  of  sides 
in  the  required  polygon ;  while  the  interior  angle,  or 
angle  between  any  two  adjacent  sides  on  the  inside  of 
the  polygon,  as  E  B  A  in  the  same  diagrams,  is  the 
supplement  of  that  angle,  or,  in  other  words,  is  found 
by  subtracting  the  exterior  angle  from  180  degrees. 
Thus  to  find  the  exterior  angle  by  means  of  which  to 
construct  a  regular  decagon,  divide  360  degrees  by  10, 
which  gives  36  degrees;  while  the  interior  angle  is 
equal  to  180  degrees  less  36  degrees,  which  is  144  de- 
grees. 


THE    ELLIPSE. 


For  a  definition  of  the  ellipse  the  reader  is  referred 
to  Chapter  I,  definitions  78  and  113.  It  may  also  be 
described  as  a  curve  drawn  with  a  constantly  increas- 
ing or  diminishing  radius,  or  as  similar  to  a  circle,  but 
having  one  diameter  longer  than  another,  the  diameters 
referred  to  being  at  right  angles  to  each  other. 


If,  upon  one  of  two  lines  intersecting  each  other 
at  right  angles,  half  of  the  long  diameter  be  set  off  each 
way  from  their  intersection  (A  A,  Fig.  206)  and  upon 
the  other  line  half  of  the  sh'ort  diameter  be  set  off  each 
way  from  the  intersection  (B  B,  Fig.  206),  four  prin- 
cipal points  in  the  circumference  of  the  ellipse  will  thus 


60 


T/ie  New  Metal    Worker  Pattern  Book. 


be  established;  and  through  these  four  points  only 
one  perfect  ellipse  can  be  drawn,  one-quarter  of  which 
is  shown  by  the  solid  line  from  A  to  B  in  the  illustra- 
tion. It  is  true  that  other  curves  having  the  appear- 
ance of  an  ellipse  can  be  drawn  through  these  points, 
as  shown  by  the  dotted  lines,  but,  as  stated  above, 
there  is  only  one  curved  line  existing  between  those 
points  which  can  be  correctly  termed  an  ellipse. 

There  are  several  methods  of  producing  a  correct 
ellipse,  as  by  a  string  and  pencil,  by  a  trammel  con- 
structed for  the  purpose  and  by  projection  from  an 
oblique  section  of  a  cylinder  or  of  a  cone,  each  of 
which  will  be  considered  in  turn.  The  ellipse  is 
properly  generated  from  two  points  upon  its  major 
axis,  called  the  foci,  and  its  circumference  is  so  drawn 
that  if  from  any  point  therein  two  lines  be  drawn  to 
the  two  foci,  their  sum  shall  be  equal  to  the  sum  of 


Fig.  206.— Defining  an  Ellipse. 

two  lines  drawn  from  any  other  point  in  the  circum- 
ference to  the  foci. 

65.  To  Draw  an  Ellipse  to  Specified  Dimensions 
with  a  String  and  Pencil.— In  Fig.  207,  let  it  be  re- 
quired to  draw  an  ellipse,  the  length  of  which  shall  be 
equal  to  the  line  A  B,  and  the  width  of  which  shall 
be  equal  to  the  line  D  C.  Lay  off  A  B  and  D  C  at 
right  angles  to  each  other,,  intersecting  at  their  middle 
points,  as  shown  at  E.  Set  the  compasses  to  one- 
half  the  length  of  the  required  figure,  as  A  E,  and 
from  either  D  or  C  as  center,  strike  an  arc,  cutting 
A  B  in  the  points  F  and  G.  These  points,  F  and  G, 
then  are  the  two  foci,  into  which  drive  pins,  as  shown. 
Drive  a  third  pin  at  C.  Then  pass  the  string  around 
the  three  points  F,  G  and  C  and  tie  it.  Eemove  the 
pin  C  and  substituting  for  it  a  pencil,  pass  the  same 
around,  as  shown  at  P,  keeping  the  string  taut.  If 
the  combined  lengths  from  F  and  G  to  the  several 
points  in  the  boundary  line  be  set  off  upon  a  straight 


line,  their  sums  will  be  found  equal.  For  example, 
the  sums  of  P  F  and  P  G,  A  F  and  A  G,  C  F  and 
C  G,  B  F  and  B  G,  are  all  the  same. 

Although  correct  so  far  as  theory  is  concerned, 
this  method  is  liable  to  error  on  account  of  the  stretch- 
ing of  the  string.  The  same  result  can  be  obtained 
by  means  of  a  trammel  constructed  for  the  purpose, 
which  is  shown  in  Fig.  208.  E  is  a  section  through 
the  arms,  showing  the  groove  in  which  the  heads  of  the 
bolts  move.  H  and  G  are  the  bolts  or  pins  by  which 
the  movement  is  controlled  and  regulated.  In  the 
engraving  the  bar  K  is  shown  with  holes  at  fixed  dis- 
tances, through  which  the  governing  pins  are  passed. 
An  improvement  upon  this  plan  of  construction  con- 
sists of  a  device  that  will  clamp  the  pins  firmly  to  the 
bar  at  any  point,  thus  providing  for  an  adjustment  of 
the  most  minute  variations. 


Fig.  SOT. — To  Draw  an  Ellipse  by  Means  of  a  String  and  Pencil. 

66.  To  Draw  an  Ellipse  to  Given  Dimensions  by  Means 
of  a  Trammel. — In  Fig.  208,  let  it  be  required  to  de- 
scribe an  ellipse,  the  length  of  which  shall  be  equal  to 
A  B  and  the  breadth  of  which  shall  be  C  D.  Draw  A 
B  and  C  D  at  right  angles,  intersecting  at  their  mid- 
dle points.  Place  the  trammel,  as  shown  in  the  en- 
graving, so  that  the  center  of  the  arms  shall  come 
directly  over  the  lines.  First  place  the  rod  along  the 
line  A  B,  so  that  the  pencil  or  point  I  shall  coincide 
with  either  A  or  B.  Then  place  the  pin  G  directly 
over  the  intersection  of  A  B  and  C  D.  Next  place 
the  rod  along  the  line  C  D,  bringing  the  pencil  or 
point  I  to  either  C  or  D,  and  put  the  pin  H  over  the 
intersection  of  A  B  and  C  D.  The  instrument  is  then 
ready  for  use,  and  the  curve  is  described  by  the  pencil 
I  moved  by  the  hand,  and  controlled  by  the  pins  work- 
ing in  the  grooves. 

When  a  trammel  is  not  convenient,  a  very  fair 
substitute  is  afforded  by  the  use  of  a  common  steel 


Geometrical    Problems. 


Gl 


square  and  a  thin  strip  of  wood,  like  a  lath.  This 
method  of  drawing  an  ellipse  is  useful  under  ordinary 
circumstances  when  only  a  part  of  the  figure  is  re- 
quired, as  in  the  shape  of  the  top  of  a  window  frame 
to  which  a  cap  is  to  be  fitted,  in  which  half  of  the 
figure  would  be  employed,  or  in  the  shaping  of  a 
member  of  a  molding  in  which  a  quarter,  or  less  than 
a  quarter,  of  the  figure  would  .be  used. 

67.  To  Draw  an  Ellipse  of  Given  Dimensions  by 
Means  of  a  Square  and  a  Strip  of  Wood.— In  Fig.  209, 
set  off  the  length  of  the  figure,  and  at  right  angles  to 
it,  through  its  middle  point,  draw  a  line  representing 
the  width  of  the  figure.  Place  a  square,  as  shown  by 
AEG,  its  inner  edge  corresponding  to  the  lines.  Lay 
the  strip  of  wood  as  shown  by  F  E,  putting  a  pencil  at 
the  point  F,  corresponding  to  one  end  of  the  figure, 


placed  upon  a  board,  and  a  line  drawn  around  it,  the 
resulting  figure  will  be  a  circle.  If  now  the  pipe  be  cut 
obliquely,  as  in  making  an  elbow  at  any  angle,  and  the 
end  thus  cut  be  placed  upon  a  board  and  a  line  drawn 
around  it,  as  mentioned  in  the  first  case,  the  figure 
drawn  will  be  an  ellipse.  What  has  thus  been  roughly 
done  by  mechanical  means  may  be  also  accomplished 
upon  the  drawing  board  in  a  very  simple  and  ex- 
peditious manner.  The  demonstration  which  follows 
is  of  especial  interest  to  the  pattern  cutter,  because 
the  principles  involved  in  it  lie  at  the  root  of  many 
practical  operations  which  he  is  called  upon  to  perform. 
For  example,  the  shape  to  cut  a  piece  to  stop  up  the 
end  of  a  pipe  or'tube  which  is  not  cut  square  across, 
the  shape  to  cut  a  flange  to  fit  a  pipe  passing  through 
the  slope  of  a  roof,  and  other  similar  requirements  of 


Fig.   SOS.— To   Draw   an   Ellipse   by 
Means  of  a  Trammel. 


Fig.  209.  Fig.  210. 

To  Draw  an  Ellipse  by  Means  of  a  Square  and  a  Strip  of  Wood. 


and  a  pin  at  E,  corresponding  to  the  inner  angle  of  the 
square.  Then  place  the  stick  across  the  figure,  as 
shown  in  Fig.  210,  making  the  pencil,  F,  correspond 
with  one  side  of  the  figure,  and  put  a  pin  at  G,  corre- 
sponding with  the  inner  angle  of  the  square.  Now 
move  the  stick  from  one  position  to  the  other,  letting 
the  points  E  and  G  slide,  one  against  the  tongue  and 
the  other  against  the  blade  of  the  square.  The  pencil 
point  will  then  describe  the  required  curve.  In  draw- 
ing the  figure  the  square  must  be  changed  in  position 
for  each  quarter  of  the  curve.  As  shown  in  the  en- 
gravings, it  is  correct  for  the  quarter  of  the  curve  rep- 
resented by  F  D,  Fig.  209.  It  must  be  changed  for 
each  of  the  other  sections,  its  inner  edge  being  brought 
against  the  lines  each  time,  as  shown. 

One  definition  of  an  ellipse  is  "  a  figure  bounded 
by  a  regular  curve,  which  corresponds  to  an  oblique 
section  of  a  cylinder." 

This  can  be  practically  illustrated  by  assuming  a 
piece  of  stove  pipe  as  tlie  representative  of  the  cylin- 
der. If  the  piece  of  pipe  is  cut  square  across,  the  end 


almost   daily   occurrence,    depend   entirely  upon   the 
principles  here  explained. 

68.  To  Describe  the  Form  or  Shape  of  an  Oblique 
Secti  n  of  a  Cylinder,  or  to  Draw  an  Ellipse  as  the 
Oblique  Projection  of  a  Circle.— The  two  propositions 
which  are  stated  above  are  virtually  one  and  the  same 
so  far  as  concers  the  pattern  cutter,  and  they  may  be 
made  quite  the  same  so  far  as  a  demonstration  is  con- 
cerned. The  explanation  of  the  engraving  is  confined 
to  the  idea  of  the  cylinder,  believing  it  in  that  shape 
to  be  of  more  practical  service  to  the  readers  of  this 
book  than  in  any  other.  In  Fig.  211,  let  G  E  F  H 
represent  any  cylinder,  and  A  B  C  D  the  plan  of  the 
same.  Let  I  K  represent  the  plane  of  any  oblique  cut 
to  be  made  through  the  cylinder.  It  is  required  to 
draw  the  shape  of  the  section  as  it  would  appear  if 
the  cylinder  were  cut  in  two  by  the  plane  I  K,  and 
either  piece  placed  with  the  end  I  K  flat  upon  paper 
and  a  line  scribed  around  it.  Divide  one-half  of 
the  plan  ABC  into  any  convenient  number  of  equal 
parts,  as  shown  by  the  figures  1,  2,  3,  4,  etc.  Through 


Tin'    \t:ir  Metal    \Vbr/,-i.  r 


litjok. 


these  points  and  tit  right  angles  to  the  diameter  A 
C  draw  lines  as  shown,  cutting  the  opposite  side  of 
the  circle.  Also  continue  these  lines  upward  until 
they  cut  the  oblique  line  I  K,  as  shown  by  I1,  2',  3', 
etc.  Draw  I'  K',  making  it  parallel  to  I  K  for  con- 
venience in  transferring  spaces.  With  the  J-square 
set  at  right  angles  to  I  K,  and  brought  successively 
against  the  points  in  it,  draw  lines  through  I'  K1, 
as  stown  by  1",  2",  3a,  etc.  With  the  dividers  take 
the  distance  across  the  plan  A  B  C  1)  on  each  of 


Fig.  211.— The  Ellipse  as  on  Oblique  Section  of  a  Cylinder. 

the  several  lines  drawn  through  it,  and  set  the  same 
distance  off  on  corresponding  lines  drawn  through 
I1  K1.  In  other  words,  taking  A  C  as  the  base  for 
measurement  in  the  one  case  and  I1  K1  as  the  base  of 
measurement  in  the  other,  set  off  from  the  latter,  on 
each  side,  the  same  length  as  the  several  lines  measure 
on  each  side  of  A  C.  Make  2"  equal  to  2,  and  3" 
equal  to  3,  and  so  on.  Through  the  points  thus  ob- 
tained trace  a  line,  as  shown  by  I'  M  K1  and  the 
opposite  side,  thus  completing  the  figure. 

To  make  this  problem  of  practical  use  it  is  neces- 
sary that  the  diameter  of  the  cylinder  shall  be  equal 
to  the  short  diameter  of  the  required  ellipse,  and  that 


UK-  lino  1  K  be  drawn  at  such  an   angle    that    the   dis- 
tance 1'  'J'  shall  be  equal  to  its  long  diameter. 

Another  definition  of  the  ellipse  is  that  "it  is  a 
liinire  bounded  by  a  regular  curve,  corresponding  to 
an  oblique  section  of  a  cone  through  its  opposite  sides." 
It  is  this  definition  of  the  ellipse  that  classes  it  among 
what  are  known  as  conic  sections.  It  is  generally  a 
matter  of  surprise  to  students  to  find  that  an  oblique 
section  of  a  cylinder,  and  an  oblique  section  of  a  cone 
through  its  opposite  sides,  produce  the  same  figure, 
but  such  is  the  case.  The  method  of  drawing  an 
ellipse  upon  this  definition  of  it  is  given  in  the  follow- 


H 


H 


Fig.  tlS.—The  Ellipse  as  an  Oblique  Section  of  a  Cone. 

ing  demonstration.  The  principles  upon  which  this 
rule  is  based,  no  less  than  those  referred  to  in  the  last 
demonstration,  are  of  especial  interest  to  the  pattern 
cutter,  because  so  many  of  the  shapes  with  which  he 
has  to  deal  owe  their  origin  to  the  cone. 

69.  To  Describe  the  Shape  of  an  Oblique  Section  of  a 
Cone  through  its  Opposite  Sides,  or  to  Draw  an  Ellipse  as 
a  Section  of  a  Cone.— In  Fig.  212,  let  B  A  C  represent 
a  cone,  of  which  E  D  G  F  is  the  plan  at  the  base. 
Let  H  I  represent  any  oblique  cut  through  its  opposite 
sides.  Then  it  is  required  to  draw  the  shape  of  tin- 
section  represented  l>v  II  I,  which  will  be  an  ellipse. 
At  any  convenient  place  outside  of  the  figure  draw  a 
duplicate  of  IF  I  parallel  to  it,  upon  which  to  construct 
the  figure  sought,  as  II'  I1.  Divide  one-half  of  the; 


Geometrica  / 


plan,  as  E  D  G,  into  any  convenient  number  of  equal 
parts,  as  shown  by  1,  2,  3,  4,  etc.  From  the  center 
of  the  plan  M  draw  radial  lines  to  these  points.  From 
each  of  the  points  also  erect  a  perpendicular  line,  which 
produce  until  it  cuts  the  base  line  B  C  of  the  cone. 
From  the  base  lino  of  the  cone  continue  each  of  these 
lines  toward  the  apex  A,  cutting  the  oblique  line  H  I. 
Through  the  points  thus  obtained  in  H  I,  and  at  right 
angles  to  the  axis  A.D  of  the  cone,  draw  lines,  as 
shown  by  1',  2',  3',  4',  etc.,  cutting  the  opposite  sides 
of  the  cone.  From  the  same  points  in  H  I,  at  right 
angles  to  it,  draw  lines  cutting  II'  I1,  as  shown  by 
r,  2",  3a,  4=,  etc.,  thus  transferring  to  it  the  same 
divisions  as  have  been  given  to  other  parts  of  the  fig- 
ure. After  having  obtained  these  several  sets  of  lines, 
the  first  step  is  to  obtain  a  plan  view  of  the  oblique 
cut,  for  which  proceed  as  follows :  With  the  di- 
viders take  -the  distance  from  the  axial  line  A  D  to 
one  side  of  the  cone,  on  each  of  the  lines  I1,  21,  3',  4', 
etc.,  and  set  off  like  distance  from  the  center  of  the 
plan  M  on  the  corresponding  radial  lines  1,  2,  3,  4, 
etc.  A  line  traced  through  the  points  thus  obtained 
will  give  the  plan  view  of  the  oblique  cut,  as  shown  by 
the  inner  line  in  the  plan. 

This  result  may  be  verified  by  dropping  lines 
vertically  from  the  points  in  H  I  across  the  plan,  in- 
tersecting them  with  the  radial  lines  in  the  plan  of 
corresponding  number.  Thus  a  line  dropped  from 
point  4  on  H  I  should  intersect  the  radial  line  M  4  at 


K -s-G— -sB 


measurement,  with  the  dividers  take  the  distance  on 
each  of  the  several  cross  lines  2J,  3s,  4s,  53,  etc.,  from 
E  G  to  one  side  of  the  plan  of  the  obi:'que  cut  just  de- 
scribed, and  set  off  the  same  distance  on  each  side  of 
II'  I1  on  the  corresponding  lines.  A  line  traced 
through  the  points  thus  obtained  will  be  an  ellipse. 


'/7r  XV"5  ,    -    .  - 

'  V^-" 

\   Vs-     7~~ 

\Ax 

^   \s'^-.. 


Fig.  SIS.-  To  Construct  an  Ellipse  from  Two  Circles  by 
Intersecting  Lines. 

the  same  point  (4s)  established  upon  it  by  measuring 
the  distance  upon  line  4'  from  A  D  to  A  B.  Having 
thus  obtained  the  shape  of  the  oblique  cut  as  it  would 
appear  in  plan,  the  next  step  is  to  set  off  upon  the 
lines  previously  drawn  through  II'  I'  the  width  of  the 
oblique  cut  in  plan  as  measured  upon  lines  of  corre- 
sponding number.  Therefore,  with  E  G  as  a  basis  of  j 


Fig.  214.— To  Draw  an  Ellipse  within  a  Oiven  Rectangle  by 
Means  of  Intersecting  Lines. 

70.  To  Construct  an  Ellipse  to  Given  Dimensions  by 
the  Use  of  Two  Circles  and  Intersecting  Lines.— In  Fig. 
213,  let  it  be  required  to  construct  an  ellipse,  the 
length  of  which  shall  equal  A  B  and  the  width  of 
which  shall  equal  H  F.  Draw  A  B  and  H  F  at  right 
angles,  intersecting  at  their  middle  points,  K.  From 
K  as  center,  and  with  one-half  of  the  length  A  B  as 
radius,  describe  the  circle  A  C  B  D.  From  K  as 
center,  and  with  one-half  of  the  width  II  F  as  radius, 
describe  the  circle  E  F  G  II.  Divide  the  larger  circle 
into  any  convenient  number  of  equal  parts,  as  shown 
bv  the  small  figures  1,  2,  3,  4,  etc.  Divide  the  smaller 
circle  into  the  same  number  of  equal  and  correspond- 
ing parts,  as  also  shown  by  figures.  By  means  of  the 
T-square,  from  the  points  in  the  outer  circle  draw 
vertical  lines,  and  from  points  in  the  inner  circle  draw 
horizontal  lines,  as  shown,  producing  them  until  they 
intersect  the  lines  first  drawn.  A  line  traced  through 
these  points  of  intersection  will  be  an  ellipse. 

11.  To  Draw  an  Ellipse  within  a  Given  Rectangle  by 
Means  of  Intersecting  Lines.— In  Fig.  214,  let  E  D  B 
A  be  any  rectangle  within  which  it  is  required  to  con- 
struct an  ellipse.  Bisect  the  end  A  E,  obtaining  the 
point  F,  from  which  erect  the  perpendicular  F  G, 
dividing  the  rectangle  horizontally  into  two  equal  por- 
tions. Bisect  the  side  A  B,  obtaining  the  point  H, 
and  draw  the  perpendicular  II  I,  dividing  the  rectangle 
vertically  into  two  equal  portions.  The  lines  F  G  and 
II  I  are  then  the  axes  of  the  ellipse.  F  G  represents 
the  major  axis,  and  II  I  the  minor  axis.  Divide  the 
spaces  F  E,  F  A,  G  D  and  G  B  into  any  convenient 
number  of  equal  parts,  as  shown  by  the  figures  1,  2,  3. 
From  these  points  in  V  K  and  G  1)  draw  lines  to  I,  and 
from  the  points  in  F  A  and  G  B  draw  lines  to  the 


11, >    New    M< />//.     \Vn,-ki'r    I'lttfcnt    Book. 


point  H.  Divide  F  C  and  G  C  also  into  the  same 
number  of  equal  parts,  as  shown  by  the  figures,  and 
from  H  and  I  through  each  of  these  points  draw  lines, 
continuing  them  till  they  intersect  lines  of  correspond- 
ing number  in  the  other  set,  as  indicated.  A  line 
traced  through  the  several  points  of  intersection  be- 
tween the  two  sets  of  lines,  as  shown  in  the  engraving, 
will  be  an  ellipse. 

Besides  the  above  methods  for  drawing  correct 
ellipses  there  are  several  methods  for  drawing  figures 
approximating  ellipses  more  or  less  closely,  but  com- 


Fig.  215.-Firat  Method. 


Fig.  216.-Second  Method. 

To  Draw  an  Apprcximate  Ellipse  with,  the  Campasteg,  the 
Length  only  Being  Given. 


posed  of  arcs  of  circles,  which  it  is  sometimes  necessary 
to  substitute  for  true  ellipses  for  constructive  reasons. 
The  ellipse  has  been  described  above  as  a  curve  drawn 
with  a  constantly  changing  radius.  If,  instead  of 
using  an  infinite  number  of  radii,  some  finite  number 
be  assumed,  it  will  appear  that  the  greater  the  number 
assumed  the  more  nearly  will  it  approach  a  perfect 
ellipse.  Thus,  a  curve  very  much  like  an  ellipse  can 
be  drawn,  each  quarter  of  which  is  composed  of  arcs 
drawn  from  two  centers.  If  the  number  of  centers  be 
increased  to  three,  the  curve  comes  much  nearer  a  true 


ellipse,  and  with  four  or  five  centers  to  each  quarter, 
the  curve  thus  produced  can  scarcely  be  distinguished 
from  the  perfect  ellipse. 

72.  To  Draw  an  Approximate  Ellipse  with  the  Com- 
passes, the  Length  only  being  Given.— In  Fig.   215,  let 
A  0  be  any  lepgth  to  which  it  is  desired  to  draw  an 
elliptical  figure.     Divide  A  C  into  four  equal  parts. 
From  3  as  center,  and  with  3  1  as  radius,  strike  the  arc 
BID,  and  from    1    as   center,  and   the  same   radius, 
strike  the  arc  B  3  D,  intersecting  the  arc  first  struck 
in  the  points  B  and  D.     From  B,  through  the  points  1 
and  3,  draw  the  lines  B  E  and  B  F  indefinitely,  and 
from  D,  in  like  manner,  draw  the  lines  D  G  and  D  II. 
From  the  point  1  as  center,  and  with   1  A   as  radius, 
strike  the  arc  E  G,  and  from  3  as  center,  with  the  same 
radius,    or  its  equivalent,    3  C,   strike  the  arc   II   F. 
From  D  as  center,  with  radius  D  G,  strike  the  arc  G  II, 
and  from  B  as  center,  with  the    same  radius,  or  its 
equivalent,  B  E,  strike  the  arc  E  F,  thus  completing 
the  figure. 

A  figure  of  different  proportions  may  be  drawn  in 
the  same  general  manner  as  follows  :  Divide  the  length 
A  C  into  four  equal  parts,  as  indicated  in  Fig.  216. 
From  2  as  center,  and  with  2  1  as  radius,  strike  the 
circle  1  E  3  F.  Bisect  the  given  length  A  C  by  the 
line  B  D,  as  shown,  cutting  the  circle  in  the  points  E 
and  F.  From  E,  through  the  points  1  and  3,  draw  the 
lines  E  G  and  E  H  indefinitely,  and  from  F,  through 
the  same  points,  draw  similar  lines,  F  I  and  F  K. 
From  1  as  center,  and  with  1  A  as  radius,  strike  the 
arc  I  A  G,  and  from  3  as  center,  with  equal  radius, 
strike  the  arc  K  C  H.  From  E  as  center,  and  with 
radius  E  G,  strike  the  arc  G  D  H,  and  from  F  as  center, 
with  corresponding  radius,  strike  the  arc  I  B  K,  thus 
completing  the  figure. 

73.  To  Draw  an  Approximate  Ellipse  with  the  Com- 
passes to  Given  Dimensions,  Using  Two  Sets  of  Centers.— 
First  Method. — In   Fig.    217,   let  A   B   represent  the 
length    of    the  required   figure  and    D  E    its    width. 
Draw  A  B  and  D  E  at  right  angles  to  each  other,  and 
intersecting  at  their  middle  points.     At  the  point  A 
erect  the  perpendicular  A  F,  and  in  length  make  it 
equal  to  C  D.     Bisect  A  F,  obtaining  the  point  N. 
Draw  N  D.     From  F  draw  a  line  to  E,  as  shown,  cut- 
ting N  D  in  the  point  G.     Bisect  the  line  G  D  by  the 
line  H  I,  perpendicular  to  G  D  and  meeting  D  E  in  the 
point  I.      In  the  same  manner  draw   lines   correspond- 
ino'  to  G  I,  as  shown  by  L  I,  M  O  and   li  0.      From  I 
and  0  as  centers,  and  with  I  G  as  radius,  strike  the 
arcs  G  D  L  and  M  E  R,  and  from  K  and  P  as  centers, 


Problems, 


65 


with  K  (!  as  radius,  strike  tin- arcs  <i  A  M  ami  L  1>  R, 
thus  completing  the  liguiv. 

74.  To  Draw  an  Approximate  Ellipse  with  the  Com- 
passes to  Given  Dimensions,  Using;  two  Sets  of  Centers. 

—Second  Method.  —  In  Fig.  21s,  lot  C  D  represent  the 
length  of  a  ro<|iiiro<l  ellipse  and  A  P>  the  width.  Lay 
oil'  these  two  dimensions  at  right  angles  to  (>aeh  other, 
as  shown.  On  C  D  lay  off  a  space  ei|iial  to  the  width 
of  the  required  figure,  as  shown  by  1)  E.  Divide  the 
remainder  of  I)  C,  or  the  space  V.  (',  into  three  equal 
parts,  as  shown  in  the  cut.  With  a  radius  equal  to 
t\vo  of  these  parts,  and  from  \\  as  center,  strike  the 
circle  G  S  F  T.  Then  with  V  and  G  as  centers,  and  V  G 
as  radius,  strike  the  arcs,  as  shown,  intersecting  upon 
A  1?  prolonged  at  ()  and  P.  From  0,  through  the 
points  G  ami  F,  draw  O  L  and  0  M,  and  likewise  from 


(•enter,  with  F  L  as  radius,  describe  a  circle,  as  shown, 
thus  establishing  the  points  M,  N  and  O,  which,  with 
L,  are  the  centers  from  which  the  ellipse  is  to  be  struck. 
From  M,  draw  M  L  Q  and  M  N  S  indefinitely,  and  in 
a  similar  manner  ()  L  P  and  0  N  R.  With  0  as  center, 
and  O  D  as  radius,  strike  the  arc  P  D  R,  cutting  O  P 
and  O  R,  as  shown.  In  a  similar  manner,  and  with 
the  same  radius  (or  which  is  the  same,  with  M  E  as 
radius)  and  M  as  center,  describe  the  arc  Q  E  S. 
With  L  and  N  as  centers,  and  with  L  15  or  N  C  as 
radius,  strike  the  arcs  Q  B  P  and  R  C  S,  thus  com- 
pleting the  figure. 

The  above  methods  of  drawing  approximate  el- 
lipses are  only  available  within  certain  limits  of  pro- 
portion, as  will  be  discovered  if  an  attempt  is  made 
to  draw  them  very  much  donga-ted,  the  limit  being 


e~  f  E 

Fig.  217. -First  Method.  Fig.  218.— Second  Method.  Fig.  219.— Third  Method. 

To  Draw  an  Approximate  Ellipse  with  the  Compasses,   Using  Two  Sets  of  Centers. 


P,  through  the  same  points,  draw  P  K  and  P  N. 
From  O  as  center,  with  O  A  as  radius,  strike  the  arc 
L  M,  and  with  the  same  radius,  and  P  as  center,  strike 
the  arc  K  N.  From  F  and  G  as  centers,  and  with  F 
D  and  G  C  as  radii,  strike  the  arcs  N  M  and  K  L  re- 
spectively, thus  completing  the  figure. 

15.  To  Draw  an  Approximate  Ellipse  with  the  Com- 
passes to  Given  Dimensions,  Using  Two  Sets  of  Centers.— 
Third  Method. — In  Fig.  219,  let  B  C  represent  the  length 
of  the  required  figure  and  D  E  its  width.  B  C  and  D  E 
are  drawn  at  right  angles  to  each  other,  intersecting  at 
their  middle  points  at  F.  The  next  step  in  describing 
the  figure  is  to  obtain  the  di (Terence  in  length  between 
the  axes  F  D  and  F  B,  which  can  be  done  as  indicated 
by  the  arc  D  G.  This  difference,  G  B,  is  to  be  set  off 
('rom  the  center  F  on  F  B  and  F  D,  as  shown  by  F  H, 
F.  J,  then  draw  II  J  and  set  off  half  of  H  J  to  L,  as 
indicated  by  the  arc  K  L.  The  object  of  the  operation 
so  far  has  been  to  secure  the  point  L.  From  F  as 


reached  when  the  long  diameter  is  about  equal  to  two 
times  the  shorter  diameter.  Beyond  this  limit  in  the 
first  two  methods,  if  the  final  arc  be  drawn  with  the 
radius  G  K  (Figs.  217  and  218),  it  will  not  reach  the 
end  of  the  long  diameter,  but  will  strike  it  at  a  point 
inside  of  A  or  C.  By  the  third  method,  if  the  long 
diameter  be  increased  until  it  is  about  2f  times  the 
shorter,  the  point  L  (Fig.  219)  will  fall  at  the  extreme 
limit  of  the  long  diameter  (B),  thus  completely  cutting 
out  the  small  arc  P  Q.  It  must,  therefore,  in  extreme 
cases  be  left  to  the  judgment  of  the  draftsman  to 
adjust  or  vary  the  lengths  of  the  radii  of  the  two  arcs 
so  as  to  produce  the  result  which  will  look  the  best. 

76.  To  Draw  an  Approximate  Ellipse  with  the  Com- 
passes to  Given  Dimensions,  Using:  Three  Sets  of  Centers. 
—In  Fig.  220,  let  A  B  represent  the  length  of  the  re- 
quired figure  and  D  E  the  width.  Draw  A  B  and  D 
E  at  right  angles  to  each  other,  intersecting  at  their 
middle  points,  as  shown  at  C.  From  the  point  A  draw 


66 


The  Xew  Metal    Worker  Pattern 


A  Fv perpendicular  to  A  B,  and  in  length  equal  to  C 
D.  Join  the  points  F  and  D,  as  shown.  Divide  A  F 
into  three  equal  parts,  thus  obtaining  the  points  Z  and 
I,  and  draw  the  lines  Z  D  and  I  D.  Divide  A  C  into 
three  equal  parts,  as  shown  by  Y  and  G,  and  draw  E 
G  and  E  Y,  prolonging  them  until  they  intersect  with 
Z  D  and  I  D  respectively,  in  the  points  J  and  II. 
Bisect  J  D,  and  draw  K  L  perpendicular  to  its  central 
point,  intersecting  D  E  prolonged  in  the  point  L.  Draw 
J  L  and  H  J.  Bisect  H  J,  and  draw  M  N  perpen- 
dicular to  its  central  point,  meeting  J  L  in  N.  Draw 
N  II,  cutting  A  B  in  the  point  0.  L  then  is  the 
center  of  the  arc  J  D  P,  N  is  the  center  of  the  arc  II 
J,  and  0  is  the  center  of  the  arc  H  A  K.  The  points 
S  and  U,  corresponding  to  N  and  0,  from  which  to 


of  X  (),  draw  P  R,  perpendicular  to  X  O  and  parallel 
to  K  M.  Then  N  (.)  and  P  K  are  tho  axes  of  the 
ellipse. 

78.  In  a  Given  Ellipse,  to  Find  Centers  by  which  an 
Approximate  Figure  may  be  Constructed — In  Fig.  -2-2-2, 
let  A  E  B  D  be  any  ellipse,  in  which  it  is  required  to 
find  centers  by  which  an  approximate  Jigure  may  be 
drawn  with  the  compasses.  Draw  the  axes  A  B  and 
E  D.  From  the  point  A  draw  A  F,  perpendicular  to 
A  B,  and  make  it  equal  to  C  E.  Join  F  and  E. 
Divide  A  F  into  as  many  equal  parts  as  it  is  desired 
to  have  sets  of  centers  for  the  figure.  In  this  in- 
stance four.  Therefore,  A  F  is  divided  into  four 
equal  parts,  as  shown  by  P<)  and  (i.  Divide  A  (' 
into  the  same  number  of  equal  parts,  as  shown  by  R 


BP 


Fig.  220. — To  Draw  an  Approximate 
Ellipse  with  the  Compasses,  Using  Three 
Sets  of  Centers. 


Fiij.  %%1. — To   Find   the   True  Axes  of  a 
Given  Ellipse. 


Fit/.  222.— In  a  Given  Ellipse,  to  Find 
Centers  by  which  mi  Approximate  Fiyure 
may  be  Constructed. 


strike  the  remainder  of  the  upper  part  of  the  figure, 
may  be  obtained  by  measurement,  as  indicated.  Hav- 
ing drawn  so  much  of  the  figure  as  can  be  struck  from 
these  centers,  set  the  dividers  to  the  distance  L  P  or  L 
J,  and  placing  one  point  at  E,  the  remaining  center 
will  be  found  at  the  other  point  of  the  dividers,  in  the 
line  E  D  prolonged,  as  shown  by  X. 

11.  To  Find  the  True  Axes  of  a.  Given  Ellipse. — In 
Fig.  221,  let  N  P  0  R  be  any  ellipse,  of  which  it  is  re- 
quired to  find  the  two  axes.  Through  the  ellipse  draw 
any  lines,  A  B  and  D  E,  parallel  to  each  other.  Bisect 
these  two  lines  and  draw  F  G,  prolonging  it  until  it 
meets  the  sides  of  the  ellipse  in  the  points  II  and  I. 
Bisect  the  line  H  I,  obtaining  the  point  C.  From  C  as 
center,  with  any  convenient  radius,  describe  the  arc 
K  L  M,  cutting  the  sides  of  the  ellipse  at  the  points  K 
and  M.  Join  K  and  M  by  a  straight  line,  as  shown. 
Bisect  M  K  by  the  line  N  0,  perpendicular  to  it. 
Through  C,  which  will  also  be  found  to  be  the  center 


S  T.  From  the  points  of  division  in  A  F  draw  lines 
to  E.  From  D  draw  lines  passing  through  the  divi- 
sions in  A  C,  prolonging  them  until  they  intersect  the 
lines  drawn  from  A  F  to  E,  as  shown  by  D  U,  D  V 
and  D  W.  Draw  the  chords  U  V,  V  W  and  W  E, 
and  from  the  center  of  each  erect  a  perpendicular, 
which  prolong  until  they  intersect  as  follows :  The 
line  perpendicular  to  W  E  intersects  the  center  line 
E  D  in  the  point  D.  Now  draw  D  W  and  prolong 
the  perpendicular  to  V  W  till  it  intersects  D  W  in  K, 
and  draw  K  V.  Prolong  the  perpendicular  to  U  V 
till  it  cuts  K  V  in  L  and  draw  L  I',  cutting  A  C  in  the 
point  S.  Then  D  is  the  center  of  the  arc  E  W,  K  is 
the  center  of  the  arc  W  V,  L  is  the  center  of  the  an; 
V  U  and  S  is  the  center  of  the  arc  U  N.  By  these 
centers  it  will  lie  seen  that  one-quarter  of  the  figure 
(A  to  E)  may  be  struck.  By  measurement,  corre- 
sponding points  may  be  located  in  other  portions  of 


the  figure. 


If  correctly  done  the  points  U,  V  and  W 


dreamt  /na.tl  Problems. 


67 


will  be  found  to  fall  upon  the  ellipse.  consequently 
the  arcs  drawn  between  those  points  from  the  centers 
obtained  cannot  deviate  much  from  the  correct  ellipse. 
79.  To  Draw  the  Joint  Lines  of  an  Elliptical  Arch.— 
First  Method. — In  a  circular  arch  the  lines  representing 
the  joints  between  the  stones  forming  the  arch,  or  the 
voussoirs  as  they  are  properly  called,  are  drawn 
radially  from  the  center  of  the  semicircle  of  the  arch. 
In  an'  elliptical  arch  this  operation  is  somewhat  more 
dillicuit,  as  the  true  ellipse  possesses  no  such  single 


B  C 

Fig  223.— First  Method. 

To  Draw  the  Joint  Lines  of  an  Elliptical  Arch. 

point,  but,  instead,  two  foci,  as  has  been  explained. 
Therefore,  the  following  course  must  be  pursued : 
From  any  point  upon  the  ellipse  at  which  it  is  desired 
to  locate  a  joint,  as  A,  Fig.  223,  draw  a  line  to  each 
of  the  foci,  as  A  B  and  A  C.  Risect  the  angle  BAG 
(Prob.  12  in  this  chapter),  as  shown  at  D,  and  extend 
the  line  D  A  outside  the  ellipse,  which  will  be  the  joint 
line  required. 

80.  To  Draw  the  Joint  Lines  of  an  Elliptical  Arch.— 
Second  Method. — In  Fig.  224,  A  R  is  one-half  the  curve 
of  the  arch,  A  C  its  center  line  and  C  B  its  springing 


line.  Draw  A  I)  parallel  to  C  B,  and  D  B  parallel  to 
A  C,  and  draw  the  diagonals  A  B  and  C  D.  From 
each  of  the  points  1,  2,  3,  etc.,  representing  the  joints, 
drop  lines  vertically,  cutting  C  D.  From  their  inter- 
sections with  C  D  carry  them  at  right  angles  to  A  B, 
cutting  the  springing  line  C  B,  as  shown  by  the  small 
figures  1",  2",  3%  etc.  From  the  points  in  C  B  draw 


3'        4"      5'  B 

Fig.  224,-Second  Method. 
To  Draw  the  Joint  Lines  of  an  Elliptical  Arch. 

lines  through  corresponding  points  in  the  arch  A  B, 
as  1'  1,  2"  2,  3'  3,  etc.,  and  continue  them  through 
the  face  of  the  arch  which  will  be  the  joint  lines 
sought. 

In  the  case  of  an  elliptical  curve  made  up  of  arcs 
of  circles,  the  joint  lines  would  be  drawn  radially  from 
the  centers  of  the  arcs  in  which  they  occur. 


THE    VOLUTE. 


The  volute  is  an  architectural  figure  of  a  geo- 
metrical nature  based  upon  the  spiral,  and  is  of  quite 
frequent  occurrence  in  one  form  or  another,  conse- 
quently some  remarks  upon  the  different  methods  of 
drawing  it  will  not  be  out  of  place. 

81.  To  Draw  a  Simple  Volute.— Let  D  A,  in  Fig. 
225,  be  the  width  of  a  scroll  or  other  member  for 
which  it  is  desired  to  draw  a  volute  termination. 
Draw  the  line  D  1,  in  length  equal  to  three  times  D 
A,  as  shown  by  D  A,  A  R  and  B  1.  From  the  point 
1  draw  1  2  at  right  angles  to  D  1,  and  in  length  equal 
to  two-thirds  the  width  of  the  scroll — that  is,  to  two- 
thirds  of  D  A.  From  2  draw  the  line  2  3  perpen- 
dicular to  1  2,  and  in  length  equal  to  three-quarters  of 
1  2.  Draw  the  diagonal  line  1  3.  From  2  draw  a 
line  perpendicular  to  1  3,  as  shown  by  2  4,  indefi- 
nitely. From  3  draw  a  line  perpendicular  to  2  3,  pro- 


ducing it  until  it  cuts  the  line  2  4  in  the  point  4. 
From  4  draw  a  line  perpendicular  to  3  4,  producing 
it  until  it  meets  the  line  1  3  in  the  point  5.  In  like 
manner  draw  5  6  and  6  7.  The  points  1,  2,  3,  4, 
etc.,  thus  obtained  are  the  centers  by  which  the  curve 
of  the  volute  is  struck.  From  1  as  center,  and  with  1 
D  as  radius,  describe  the  quarter  circle  D  C.  Then 
from  2  as  center,  and  2  C  as  radius,  describe  the 
quarter  circle  C  F,  and  so  continue  using  the  centers 
in  their  numerical  order  until  the  curve  intersects  with 
the  other  curve  beginning  at  A  and  struck  from  the 
same  centers,  thus  completing  the  figure,  as  shown. 

82.  To  Draw  an  Ionic  Volute. — Draw  the  line  A 
B,  Fig.  226,  equal  to  the  hight  of  the  required  volute, 
and  divide  it  into  seven  equal  parts.  From  the  third 
division  draw  the  line  3  C,  and  from  a  point  on  this 
line  at  any  convenient  distance  from  A  B  describe  :i 


68 


The  Xcw  Mdal    Worker   I'allcnt  Book. 


circle,  the  diameter  of  which  shall  equal  one  of  the 
seven  divisions  of  the  line  A  B.  This  circle  forms 
the  eye  of  the  volute.  In  order  to  show  its  dimen- 
sions, etc.,  it  is  enlarged  in  Fig.  227.  A  square,  D 
E  F  G,  is  constructed,  and  the  diagonals  G  E  and  F 
D  are  drawn.  F  E  is  bisected  at  the  point  1,  and  the 
line  1  2  is  drawn  parallel  to  G  E.  The  line  2  3  is 
then  'drawn  indefinitely  from  2  parallel  to  F  D,  cut- 
ting G  E  in  the  point  II.  The  distance  from  H  to  the 
center  of  the  circle  0  is  divided  into  three  equal 
parts,  as  shown  by  H  a  b  O.  The  triangle  2  0  1  is 
formed.  On  the  line  0  H  set  off  a  point,  as  c,  at  a 
distance  from  0  equal  to  one-half  of  one  of  the  three 
equal  parts  into  which  0  H  has  been  divided.  From 
c  draw  the  line  c  3  parallel  to  1  O,  producing  it  until 
it  cuts  2  3  in  the  point  3.  From  3  draw  the  line  3  4 
parallel  to  G  E  indefinitely.  From  the  point  c  draw  a 
line  c  4  parallel  to  2  0,  cutting  the  line  3  4  -in  the 
point  4,  completing  the  triangle  c  3  4.  From  4  draw 
the  line  4  5  parallel  to  F  D,  meeting  1  0  in  the  point 
5.  From  5  draw  the  line  5  6  parallel  to  G  E,  meeting 
the  line  2  0  in  the  point  6.  From  6  draw  the  line,  6 
7  parallel  to  F  D,  meeting  the  line  c  3  in  the  point  7. 


Fig.  225.— To  Draw  a  Simple  Volute. 

Proceed  in  this  manner,  obtaining  the  remaining  points, 
8,  9,  10,  11  and  12.  These  points  form  the  centers 
by  which  the  outer  line  of  the  volute  proper  is  drawn. 
From  1  as  center,  and  with  radius  1  F,  Fig.  226,  de- 
scribe the  quarter  circle  F  G.  Then  from  2  as  center, 
and  with  radius  2  G  describe  the  quarter  circle  G  D, 
and  so  continue  striking  a  quarter  circle  from  each  of 
the  centers  above  described  until  the  last  arc  meets 


the  circle  first  drawn.  To  obtain  the  centers  by  which 
the  inner  line  of  the  volute  is  struck,  and  which 
gradually  approaches  the  outer  line  throughout  its 
course,  proceed  as  follows :  Produce  the  line  3  c, 
Fig.  227,  until  it  intersects  1  2  in  the  point  I1,  which 


Fig.  S26.—To  Draw  an  Ionic  Volute. 

mark.  This  operation  gives  also  the  points  ',)'  and  5* 
of  intersection  with  the  lines  parallel  to  1  2,  whidi 
also  mark.  In  like  manner  produce  4  c,  1  O  and  2 
O,  as  shown  by  the  dotted  lines,  and  mark  the  several 
points  of  intersection  formed  with  the  cross  lines. 
Then  the  points  I1,  2',  3',  4',  etc.,  thus  obtained  are 
the  centers  for  the  inner  line  of  the  volute,  which 
use  in  the  same  manner  as  described  for  producing 
the  outer  line. 

83.  To  Draw  a  Spiral  from  Centers  with  Compasses. 
— Divide  the  circumference  of  the  primary — some- 
times called  the  eye  of  the  spiral — into  any  number 
of  equal  parts;  the  larger  the  number  of  parts  the 
more  regular  will  be  the  spiral.  Fig.  228  shows  the 
primary  divided  into  six  equal  parts.  Fig.  22!»  is  an 
enlarged  view  of  this  portion  of  the  preceding  figure. 
Construct  the  polygon  by  drawing  the  lines  1  2,  2  3, 
3  4,  etc..  producing  them  outside  of  the  primary,  as 
shown  by  A,  B,  D,  F,  0  and  E.  From  2  as  center, 
with  2  1  as  radius,  describe  the  arc  A  B.  From  3  as 
center,  and  3  B  as  radius,  describe  the  arc  B  D;  and 


(li'iinn 


C,!) 


with  4  as  center,  witli  radius  4  D,  describe  the  arc  D 
F.  In  this  manner  the  spiral  may  be  continued  ;(iiv 
nuiuber  of  revolutions.  Jn  the  resulting  ligure  the 
various  revolutions  will  be  parallel. 

84.  To  Draw  a  Spiral  by  Means  of  a  Spool  and 
Thread. — Set  the  spool  as  shown  bv  A  D  B    in  Fig. 


Fig.  227.— Eye.  of  the  Volute  in  Fig.  226  Enlarged. 

230  and  wind  a  thread  around  it.  Make  a  loop,  E, 
in  the  end  of  the  thread,  in  which  plaee  a  pencil,  as 
shown.  Hold  the  spool  firmly  and  move  the  pencil 
around  it,  unwinding  the  thread.  A  curve  will  be  de- 
scribed, as  shown  in  the  dotted  lines  of  the  engrav- 
ing. It  is  evident  that  the  proportions  of  the  figure 


toj),  K  A  at  the  bottom  and  A  B  at  the  side,  the 
length  of  A  B,  which  determines  the  width  of  the 
scroll,  being  given.  Bisect  A  B,  obtaining  the 
point  C.  Let  the  distance  between  the  beginning 
and  ending  of  the  first  revolution  of  the  scroll,  shown 
by  a  c.  l)o  established  at  pleasure.  Having  determined 


Fig.  231.— To  Draw  a  Scroll  to  a 
Specified  Width. 


Fig.  233.— The  Center  of  Fig. 
SSI  Enlarged. 


this  distance,  take  one-eighth  of  it  and  set  it  off  up- 
ward from  C  on  the  line  A  B,  thus  obtaining  the  point 
b.  From  l>  draw  a  horizontal  line  of  any  convenient 
length,  as  shown  by  b  h.  With  one  point  of  the  com- 
passes set  at  b,  and  with  b  A  as  radius,  describe  an  arc 
cutting  the  line  b  h  in  the  point  1.  In  like  manner, 
from  the  same  center,  with  radius  b  B,  describe  an  arc 
cutting  the  line  b  h  in  the  point  2.  Upon  1  2  as  a  base 
erect  a  square,  as  shown  by  1  2  3  4.  Then  from  1  as 


Fig.  228.— To  Draw  a  Spiral  from  Centers. 


Fig.  129.— Enlarged  View  of  the  Eye  of  the  Spiral  in 
Fig.  228. 


Fig.    SSO.—To    Draw   a   Spiral    by 
Means  of  a  Spool  and  Thread. 


are  determined  by  the  size  of  the  spool.  Hence  a 
larger  or  smaller  spool  is  to  be  used,  as  circumstances 
require. 

85.  To  Draw  a  Scroll  to  a  Specified  Width,  as  for  a 
Bracket  or  Modillion. — In  Fig.  231,  let  it  be  required  to 
construct  a  scroll  which  shall  touch  the  line  D  B  at  the 


center,  with  1  a  as  radius,  describe  an  arc,  ab;  and 
from  2  as  center,  with  2  b  as  radius,  describe  the  arc 
b  c.  From  3  as  center,  with  radius  3  e,  describe  the 
arc  c  d.  From  4  as  center,  with  radius  4  d,  describe 
the  arc  d  e.  If  the  curve  were  continued  from  e,  being 
struck  from  the  same  centers,  it  would  run  parallel  to 


70 


Tin'    \rir    MI  ti/l     \Vnrl-fl-    1'iilli'i'n     linn/.-. 


itself;  but  as  the  inner  line  of  the  scroll  runs  parallel 
to  the  outer  line,  its  width  may  be  set  off  at  pleasure, 
as  shown  by  a  a',  and  the  inner  line  may  be  drawn  by 
the  same  centers  as  already  used  for  the  outer,  and  con- 
tinued until  it  is  intersected  by  the  outer  curve.  To 
find  the  centers  from  which  to  complete  the  outer 
curve,  construct  upon  the  line  of  the  last  radius  above 
used  (4  e)  a  smaller  square  within  the  larger  one,  as 
shown  by  5678.  Thi?  is  better  illustrated  by  the 
larger  diagram,  Fig.  232,  in  which  like  figures  repre- 
sent the  same  points.  Make  the  distance  from  5  to  8 


equal  to  one-half  of  the  space  from  4  to  1,  making 
4  to  8  equal  the  distance  of  5  to  1.  Make  5  to  6  equal 
the  distance  from  8  to  5.  After  obtaining  the  points 
5,  (!,  7,  etc.,  in  this  manner,  so  many  of  them  arc  i<> 
be  used  as  are  necessary  to  make1  the  outer  curve  inter- 
sect the  inner  one,  as  shown  at  <j.  Thus  5  is  used  as 
a  center  for  the  arc  e/,  and  (i  as  a  center  for  the  are 
f  g.  If  the  distance  a  a'  were  taken  less  than  here 
given,  it  is  easy  to  see  that  more  of  the  centers  upon 
the  small  square  would  require  to  be  used  to  arrive  at 
the  intersectiou. 


p 

x_/ 


CHAPTER  V. 


To  any  one  wishing  to  pursue  pattern  cutting  as 
a  profession  it  is  essential  not  only  that  lie  know  how 
to  solve  a  large  number  of  intricate  problems,  but  that 
he  understand  thoroughly  the  principles  which  under- 
lie  such  operations.  It  is,  therefore,  appropriate,  be- 
fore introducing  pattern  problems,  that  some  attention 
should  be  given  t<>  the?  explanation  of  such  principles 
in  order  that  the  reasons  for  the  steps  taken  in  the 
demonstrations  following  mav  be  readily  understood. 
Underlying  the  entire  range  of  problems  peculiar  to 
sheet  metal  work  are  certain  fundamental  principles, 
which,  when  thoroughly  understood,  make  plain  and 
simple  that  which  otherwise  would  appear  arbitrary, 
if  not  actually  mysterious.  So  true  is  this  that  noth- 
ing is  risked  in  asserting  that  any  one  who  thoroughly 
comprehends  all  the  steps  in  connection  with  cutting 
a  simple  square  miter  is  able  to  cut  any  miter  what- 
soever. Since  almost  any  one  can  cut  a  square  miter, 
the  question  at  once  arises,  in  view  of  this  statement, 
why  is  it  that  he  cannot  cut  a  raking  miter,  or  a  pin- 
nacle miter,  or  any  other  equally  difficult  form?  The 
answer  is,  because  he  does  not  understand  how  he  cuts 
the  square  miter.  He  may  perform  the  operation 
just  as  he  has  been  taught,  and  produce  results  entirely 
satisfactory  from  a  mechanical  standpoint,  without  be- 
ing intelligent  as  to  all  that  he  has  done.  He  does 
not  -comprehend  the  w\\y  and  wherefore  of  the  steps 
taken.  Hence  it  is  that  when  he  undertakes  some 
other  miter  he  finds  himself  deficient. 

There  is  a  wide  difference  between  the  skill  that 
produces  a  pattern  by  rote — by  a  mere  effort  of  •  the 
memorv — and  that  which  reasons  out  the  successive 
steps.  One  is  worth  but  very  little,  while  the  other 
renders  its  possessor  independent.  It  is  with  a  desire 
to  put  the  student  in  possession  of  this  latter  kind  of 
skill,  to  render  him  intelligent  as  to  every  operation 
to  be  performed,  that  the  present  chapter  is  written. 

The  forms  with  which  the  pattern  cutter  has  to 
deal  may  be  divided,  for  convenience  of  deseription, 
into  three  general  classes  : 


I.  The  first  of   these   embraces  moldings,    pipes 
and  r/'ijnhu-  continuous  forms,  and  may  be  called  forms 
of  parallel  lines,  or  as  a   shorter  and  more   convenient 
name  to  use,  parallel  forms. 

II.  The  second,  which  will  be  called  regular  taper- 
ing forms,  comprehends  all  shapes  derived  from  cones 
or  pyramids,  or  from  solids  having  any  of  the  regular 
geometric  figures  as  a  base  and  which  terminate  in  an 
apex. 

III.  The  third  class  will  be  called  irregular  forms, 
and  will  include  everything  not  classified  under  either 
of  the  two  previous  heads.     Many  of  these  might  be 
properly  called  transition  pieces— that  is,  pieces  which 
have    figures    of    various    outlines    placed    at    various 
angles  as  their  bases,  and  have  figures  with  differing 
outlines  variously  placed,  as  their  upper  terminations, 
thus  forming  transitions,  or  connecting  pieces  between 
the   form  which  lies  next   them   at  one  end  and  the 
adjacent  form  on  the  other  end. 

While  pieces  of  metal  of  any  shape  necessary  to 
form  the  covering  of  a  solid  of  any  shape  may  prop- 
erly be  called  patterns,  the  shapes  of  pieces  necessary 
to  form  the  joints  between  moldings  meeting  at  an 
angle  are  known  distinctively  as  miters.  This  name 
applies  equally  well  in  sheet  metal  work  if  the  two 
arms  of  the  moldiag  are  not  of  the  same  profile,  or  to 
a  single  arm  coming  against  any  plain  or  irregular  sur- 
face. These  forms  comprise  the  first  class  referred  to 
above  and,  so  far  as  principle  is  concerned,  come  under 
the  same  general  rules,  which  will  be  subsequently 
given. 

Conical  forms,  with  very  little  taper,  coming 
against  other  forms  are  also  said  to  miter  with  them. 
In  fact,  the  word  miter  has  come  into  such  general  use 
that  it  is  often  applied  to  any  joint  between  pieces  of 
metal ;  but  the  term  can  scarcely  be  considered  as  cor- 
rect when  the  forms  have  very  much  taper.  The 
principle  involved  in  the  development  of  such  patterns, 
however,  is  the  same  as  that  applied  to  the  develop- 
ment of  the  surfaces  of  all  other  regular  tapering  forms, 


Tin:   .\i-n-   Mitnl     \\'i>/-/,-rr    I'atl'fii     I  in, I,. 


referred  to  above  as  the  second  class,  whose  character- 
istics will  be  considered  in  their  proper  chapter. 

The  method  employed  for  developing  the  patterns 
for  forms  of  the  third  class  has  been  termed  triinii/nlii- 
tion,  and  is  adopted  on  account  of  its  simplicity,  as  it 
does  away  with  the  reduction  or  subdivision  of  an 
irregular  form  into  a  number  of  smaller  regular  forms, 
each  one  of  which  would  have  to  be  treated  separately 
and  perhaps  by  a  different  method.  In  fact,  there  are 
some  shapes  which  have  arisen  from  force  of  circum- 
stances which  it  would  be  impossible  to  separate  into 
regular  parts,  and  even  if  they  could  be  so  separated 
such  a  course  would  result  in  tedious  and  complicated 
operations. 

After  principles  have  been  thoroughly  explained 
the  problems  in  this  work  will  follow  in  three  sections 
or  departments  of  the  final  chapter,  arranged  according 
to  the  above  classification. 


This  is  one  of  the  instances  in  which  the  pattern  cut- 
ter is  required  to  be  something  of  an  architectural 
draftsman,  and  to  this  end  a  chapter  ofi  Linear  Draw- 
ing (Chap.  Ill)  has  been  introduced,  in  which  atten- 
tion is  given  to  this  phase  of  the  work,  and  to  which 
the  student  is  referred. 

The  arrangement  of  the  problems  in  each  of  the 
sections  of  the  succeeding  chapter  will  be  made  with 
reference  to  these  two  conditions,  the  simpler  ones 
being  placed  before  those  in  which  preliminary  draw- 
ing is  required. 

Parallel   Forms. 

(MITER  CUTTING.) 

Since  in  sheet  metal  work  a  molding  is  made  by 
bending  the  sheet  until  it  fits  a  given  stay,  a  molding 
may  be  defined  mechanically  as  a  succession  of  paral- 


Fig.  23S.— Profile  of  a  Molding. 


Fig.  XS4.—A  Stay. 


Fig.  235. — A  Reverse  Stay. 


Two  conditions  exist  in  regard  to  the  work  of 
developing  patterns  of  all  forms,  no  matter  to  which 
of  the  three  classes  above  defined  they  may  belong : 

yifxt — In  very  many  cases  a  simple  elevation  or  plan 
of  the  intersecting  parts,  together  with  their  profiles,  is 
all  that  is  necessary  to  begin  the  work  of  developing 
the  pattern — that  is,  the  plan  or  elevation,  as  the  case 
may  be,  shows  the  line  (either  straight  or  curved) 
which  represents  the  surface  against  which  another 
part  is  to  be  fitted ;  in  other  words,  the  much  sought 
for  "  miter  line." 

X'miul — In  numerous  other  instances,  however, 
no  view  can  be  drawn  either  in  elevation,  oblique  or 
otherwise,  or  in  plan,  in  which  the  miter  or  junction 
of  the  parts  will  appear  as  a  simple  straight  or  curved 
line  against  which  the  points  can  be  dropped.  In  such 
cases  it  becomes  necessary  to  do  some  preliminary 
work  in  order  to  prepare  the  way  to  the  actual  work 
of  laying  out  the  pattern.  A  view  of  the  joint  must 
be  developed  by  means  of  intersections  of  lines  which 
will  show  it  as  it  appears  in  connection  with  the  eleva- 
tion or  plan  to  be  used  in  developing  the  pattern. 


lei  forms  or  bends  to  a  given  stay,  and,  so  far  as  the 
mechanic  is  concerned,  any  continuous  form  or  ar- 
rangement of  parallel  continuous  forms,  made  for  any 
purpose  whatever,  may  be  considered  a  molding  and 
treated  under  the  same  rules  in  all  the  operations  of 
pattern  cutting.  Keeping  this  fact  in  mind  all  paral- 
lel forms  will  be  considered  as  moldings  and  that 
word  will  be  used  in  the  demonstrations,  remembering 
that  a  difference  in  name  simply  means  a  difference  of 
profile,  but  not  a  difference  in  treatment  or  principle. 
A  molding  may  be  defined  theoretically  as  a  form 
or  surface  generated  by  a  profile  passed  in  a  straight 
or  curved  line  from  one  point  to  another,  this  profile 
being  the  shape  that  would  be  seen  when  looking  at 
its  end  if  the  molding  were  cut  off  square.  A  prac- 
tical illustration  of  this  may  be  given  as  follows:  In 
Fig.  233,  let  the  form  shown  be  the  profile  of  some 
molding.  If  this  shape  be  cut  out  of  tin  plate  or 
sheet  iron,  as  shown  in  Fig.  234,  it  is  called  a  stay. 
For  the  purpose  of  this  illustration,  as  will  appear  fur- 
ther on,  a  stay,  the  reverse  of  the  one  shown  in  Fig. 
234,  or,  in  other  words,  the  piece  cut  from  the  face  or 


i if    Pull, -i-il     Cull! inf. 


outside   of    the   shape    represented    in    that    figure,    as 
shown  in  Fig.  •!'•'>•>,  will  !>«•  required. 

Having  made  a  reverse  stav,  or  " outside  stay, 
as  it  is  sometimes  called,  Fig.  :>:>;>,  take  some  plastic 
material — as  potters'  day — and,  placing  it  against  any 
smooth  surface,  as  of  a  hoard,  place  the  stay  against 
the  board  near  one  end  in  such  a  position  that  its  ver- 
tical lines  are  parallel  with  the  ends  of  the  board,  and 
move  this  reverse  stav  in  a  straight  line  along  the  faee 
of  the  board  until  a  continuous  form  is  obtained  in  the 
dav  corresponding  to  the  profile  of  the  stay,  all  as 
illustrated  in  Fig.  -!'!<>.  Bv  this  operation  will  be 
produced  a  molding  in  accordance  with  the  second 
definition  above  given.  The  purpose  in  introducing 
this  illustration  is  to  show  more  dearly  than  is  other- 
wise possible  the  principles  upon  which  the  different 


Firj.  236.— Generating  a  Molding  in  Plastic  Material  by  Means  of  a  Reverse  Stay 


parts  of  a  molding  are  measured  in  the  process  of  pat- 
tern cutting. 

Suppose  that  the  form  produced  as  illustrated  in 
Fig.  23 (5  be  completed,  and  that  both  ends  of  the 
molding  be  cut  off  square.  It  is  evident,  upon  in- 
spection, that  the  length  of  a  piece  of  sheet  metal 
necessary  to  form  a  covering  to  this  molding  will  be 
the  length  of  the  molding  itself,  and  that  the  width  of 
the  piece  will  be  equal  to  the  distance  obtained  In- 
measuring  around  the  outline  of  the  stay  which  was 
used  in  giving  shape  to  the  molding.  Now  with  a 
thin-bladed  knife,  or  by  means  of  a  piece  of  fine  wire 
stretched  tight,  let  one  end  of  the  clay  molding  just 
constructed  be  cut  off  at  any  angle.  By  inspection  of 
the  form  when  thus  cut,  as  clearly  shown  in  the  upper 
part  of  Fig.  237,  it  is  evident  that  the  end  of  a  pattern 
to  form  a  covering'  of  this  model  must  have  such  a 
shape  as  will  make  it  when  formed  up  conform  to  the 
oblique  end  of  the  molding  or  model. 


To  cut  such  a  pattern  by  means  of  a  straight  line 
drawn  from  a  point  corresponding  to  the  end  of  the 
longer  side  of  the  mold,  to  a  point  corresponding  to 
the  end  of  the  shorter  side  of  it,  would  not  be  right, 
evidently,  because  certain  parts  of  the  covering,  when 
formed  up,  fold  down  into  the  angles  of  the  molding, 
and  therefore  would  require  to  be  either  longer  or 
shorter,  as  the  case  might  be,  than  if  cut  as  above  de- 
scribed. It  is  plain,  then,  that  some  plan  must  be 
devised  by  which  measurements  can  be  taken  in  all 
these  angles  or  bends,  and  at  as  many  intermediate 
points  as  may  be  necessary,  in  order  to  obtain  the  right 
length  at  all  points  throughout  its  width.  This  can 
be  done  quite  simply  as  follows  : 

Divide  the  curved  parts  of  the  stay  into  any  con- 
venient number  of  equal  parts,  and  at  each  division 
cut  a  notch,  or  affix  a  point  to  it. 
Replace  the  stay  in  the  position 
it  occupied  in  producing  the 
molding  shown  in  Fig.  236  and 
pass  it  again  over  the  entire 
length  of  the  model.  The  points 
fastened  to  the  stay  will  then 
leave  tracks  or  lines  upon  the 
surface  of  the  molding.  Now, 
by  means  of  measurement  upon 
the  different  lines  thus  produced, 
the  length  of  the  molding  at  all  of 
the  several  points  established  in 
the  stay  may  be  obtained.  All 
this  is  clearly  illustrated  in  Fig. 

237.  In  the  upper  right  hand  corner  of  the  illus- 
tration is  shown  the  stay  prepared  with  points,  by 
moving  which  as  above  described  lines  are  left  upon 
the  face  of  the  molding,  as  shown  to  the  left 

Now,  upon  a  sheet  of  paper  fastened  to  a  draw- 
ing board,  draw  a  vertical  line,  as  shown  by  A  B  in 
Fig.  237,  and  upon  that  line  set  off  with  the  dividers 
the  width  of  each  space  or  part  of  the  profile  or  stay — 
that  is,  make  the  space  1  2  in  the  line  A  B  equal  to 
the  space  1  2  in  the  stay,  and  2  3  in  the  line  A  B 
equal  to  2  3  of  the  stay,  and  so  continue  until  all  the 
spaces  are  transferred — and  from  the  points  thus  ob- 
tained in  A  B  draw  lines  at  right  angles  to  it  indefi- 
nitely, as  shown  to  the  left.  The  lines  an  dspaces  upon 
the  paper  will  then  correspond  to  the  lines  and  spaces 
upon  the  clay  molding  made  by  the  points  fastened  to 
the  stay.  Next,  measure  with  the  dividers  the  length 
of  the  molding  upon  each  of  the  lines  drawn  upon  it, 
and  set  off  the  same  lengths  upon  the  corresponding 


T/IC 


l     \\'<n-l,-rr    J'iitti'i-il 


lines  drawn  upon  the  paper.  This  gives  :i  scries  of 
points  through  which  a.  line  may  lie  traced  which  will 
correspond  in  shape  to  the  oblique  end  of  the  molding. 
Thus,  set  off  from  A  B  on  the  line  1  on  the  paper  the 
length  of  the  molding,  measured  from  its  straight  end 
to  its  oblique  end,  upon  the  line  produced  by  point  1 
of  the  stav  upon  its  face:  and  upon  each  of  the  other 
lines  on  the  paper  set  off  the  length  of  the  molding  on 
the  corresponding  line  on  its  face,  meas- 
uring from  the  square  end  each  time, 
which  is  represented  by  the  line  A  B  of 
the  drawing.  By  this  means  are  ob- 
tained points  through  which,  if  a  line  be 
traced,  as  shown  by  C  D,  the  pattern  of 
the  covering  will  be  described.  The  line 
A  B,  containing  measurements  from  the 
profile,  is  called  the  "stretchout  line," 
and  the  lines  drawn  through  the  points  in 
it  and  at  right  angles  to  it  arc  mathe- 
matically known  as  ordinates,  but  will  in 
this  work  be  called  "  measuring  lines." 

Now,  what  has  been  done  in  Fig. 
237  illustrates  what  is  called  "  miter  cut- 
ting," which  in  other  words  consists  in 
describing  upon  a  flat  surface  the  shape  of 
a  given  form  or  envelope,  so  that  when 
the  envelope  is  cut  out  of  the  flat  sur- 
face and  formed  up  to  the  stay  from  which 
its  stretchout  was  derived,  the  finished 
molding  will  fit  against 'a  given  surface  at 
a  given  angle  previously  specified. 

The  pattern  shown  in  the  lower  part 
of  Fig.  237,  which  has  been  obtained  by 
means  of  a  clay  model,  and  measurements 
for  which  were  obtained  from  the  lines 
drawn  on  the  surface  of  the  clay  model — 
mav  be  obtained  just  as  well  from  a  draw- 
ing.     The  question  then  is,  how  can  the 
same  results  be  obtained  by  lines  drawn  upon    a  flat 
surface  as  were   obtained  by  measurements  on  lines 
drawn  along  the  surface  of  a  molding? 

In  moving  the  stay  along  the  clay  molding,  cer- 
tain lines  were  made  by  means  of  the  -points  affixed. 
If  the  reader  will  carefully  examine  Fig.  237  he  will 
notice  that  the  lines  upon  the  molding  made  by  this 
means  corresponded  in  number  and  position  with  the 
points  in  the  profile  when  it  is  laid  flat  on  its  side,  in  a 
position  exactly  opposite  the  end  of  the  modef,  as  shown. 

Hence,  if  the  profile  be  drawn  upon  paper  and  in 
line  with  it,  the  elevation  terminated  by  the  oblique 


line,  which  represents  the  surface  against  which  it  is 
required  to' miter,  the  same  results  can.  he  accom- 
plished, care  only  being  necessary  that  the  relative 
positions  of  the  parts  lie  correctly  maintained. 

This  is  illustrated  in  Fig.  23s,  which  is  to  be 
compared  with  Fig.  237,  and  shows  :  First,  that  the 
profile  A  is  drawn  in  correct  position.  Next,  that 
from  it  the  elevation  F  C  D  G  of  the  molding  is  pro- 


Fig.  SS7.—The  Use  of  Lines  in  Obtaining  the  Envelope  of  a  Molding  from  a  Model 

of  the  Same. 

jected,  as  follows:  I'se  the  T-square  in  the  general 
position  shown  by  B  in  the  engraving,  bringing  it 
against  the  several  points  in  A.  in  order  to  draw  the 
lines.  Draw  a  line  for  each  of  the  angles  in  the  profile 
A,  and  also  one  corresponding  to  each  of  the  inter- 
mediate points  in  the  curved  parts  of  the  stay.  Draw 
the  line  F  G,  representing  the  oblique  cut,  and  the 
line  C  D,  representing  the  straight  end.  Then  it  will 
be  seen  that  F  C  D  G  of  Fig.  238,  so  far  as  lines  are 
concerned,  is  exactly  the  same  as  the  molding  made 
of  clay,  shown  in  Fig.  237.  The  line  F  G,  bv  the 
definition  of  a  miter.,  is  the  "miter  line"  of  this 


Principles  nf 


molding.  It  represents  the  surface  against  which  the 
end  of  the  molding  is  supposed  to  lit.  Next  lay  oil'  a, 
stretchout  of  the  profile  A,  in  the  same  manner  as  de- 
scribed in  eiiiiiieetion  which  Fig.  'I'M,  all  as  shown  hy 
II  K  in  Fig.  23s,  through  the  points  in  which  draw 
measuring  lines  at  right  angles  to  it,  or,  what  is  the 
same,  parallel  to  the  lines  of  the  moldings.  Now, 
make  each  of  these  lines  equal  in  length  to  the  line  of 


Fig.  23S. — Obtaining  the  Envelope  of  a  Molding  from  a  Drawing  of  the  Same  by  the 

Use  of  the 


corresponding  number  drawn  across  the  elevation  from 
C  D  to  F  G. 

If,  as  suggested  in  the  previous  illustration — that 
is,  bv  using  a  pair  of  dividers  to  measure  the  length  of 
the  molding  from  C  D  to  F  G  on  the  several  lines — 
these  lengths  be  set  off  on  corresponding  lines  drawn 
from  the  stretchout  line  II  K,  a  pattern  will  be  ob- 
tained in  all  respects  corresponding  to  the  pattern 
shown  in  Fig.  237,  already  referred  to.  By  inspection 
of  the  result  thus  obtained,  however,  it  will  be  seen 
that  each  point  in  L  M  is  directly  under  the  point  of 
corresponding  number  in  line  F  G,  and  that  the  same 
thing  may  be  accomplished  by  using  the  T-square  placed 


in  the  position  shown  by  the  dotted  lines  in  Fig.  238. 
Therefore,  instead  of  using  the  dividers  proceed  as 
follows :  Place  the  T-square  as  shown  at  E,  and, 
bringing  it  successively  against  the  points  in  F  G,  cut 
measuring  lines  of  corresponding  number  by  means  of 
a  dot  or  short  dash  placed  across  the  line.  Then  a  line 
traced  as  before  through  the  points  of  intersection  thus 
obtained,  as  shown  from  L  to  M,  will  be  the  shape  of  the 
pattern  necessary  to  make  it  fit  against  a 
surface  placed  at  the  angle  represented 
by  the  miter  line  F  G.  By  this  illus- 
tration it  is  shown  that  the  T-scluare 
may  be  used  with  great  advantage  in 
transferring  measurements  under  almost 
all  circumstances.  Since  now  the  T- 
square  is  to  be  used  instead  of  the 
dividers  to  locate  the  points  in  the  pat- 
terns, the  stretchout  line  is  not  needed 
as  a  starting  point  from  which  to  meas- 
ure lengths  and  may,  therefore,  be 
located  at  will.  For  convenience,  it 
should  be  placed  as  near  to  the  miter 
line  as  possible.  Hence,  in  practical 
work,  supposing  that  the  molding  rep- 
resented by  F  C  D  G  is  not  a  very 
short  piece,  the  stretchout  line,  instead 
of  being  opposite  the  end  C  D,  would 
be  placed  somewhere  near  the  line  of  the 
blade  of  the  T-square  when  in  its  posi- 
tion at  E.  Should  the  arm  required  be 
short,  a  line  drawn  opposite  the  square 
end  will  serve  the  double  purpose  of  a 
stretch-out  line  and  of  the  outline  of 
the  square  end  of  the  pattern. 

By  further  inspection  of  Fig.  238, 
it  will  be  seen  that,  instead  of  drawing 
the  lines  from  the  points  in  the  profile 
A  the  entire  length  of  the  molding,  as  there  shown, 
all  that  is  necessary  to  the  operation  is  a  short  line 
corresponding  to  each  of  the  points  of  the  profile, 
extending  only  across  the  miter  line  F  G.  The  use  of 
these  lines,  it  is  evident,  is  only  to  locate  intersections 
upon  the  mitfcr  line.  In  other  words,  all  that  is  needed 
is  the  points  in  the  profile  A  transferred  to  the  miter 
line  F  G.  The  operation  of  transferring  these  points 
by  short  lines,  as  above  described,  is  termed  ' '  drop- 
ping the  points  ' '  from  the  profile  to  the  miter  line. 

If,  irAtead  of  the  molding  terminating  against  a 
plane  surface,  as  shown  by  F  G  in  Fig.  238,  it  be  re- 
quired to  develop  a  pattern  to  fit  against  an  irregular 


76 


The  New  Mckd    II 


Hook. 


surface,  the  method  of  procedure  would  be  exactly  the 
same,  simply  substituting  for  the  straight  Hue  F  G  a 
representation  of  that  surface.  From  this  it  will  In- 
seen  that  all  that  is  required  to  develop  the  pattern  of 
any  miter  is  that  a  correct  representation  (elevation  or 
plan)  of  the  molding  be  made,  showing  the  angle  of 
the  miter,  and  that  a  profile  be  so  drawn  that  it  shall 
be  in  line  with  the  elevation  of  the  molding — its  face 


fig.  239. — Comparison  Between  a  Butt  Miter  and  a  Miter  Between 
Two  Moldings  at  Any  Angle. 

being  so  placed  as  to  agree  with  the  face  of  the  mold- 
ing— and  that  points  from  the  subdivisions  of  the  pro- 
file be  carried  parallel  to  the  molding,  their  intersec- 
tions with  the  miter  line  being  marked  by  short  lines. 

In  order  to  more  clearly  indicate  the  point  desired 
by  this  summary  of  requirements,  suppose  that  upon 
each  of  two  pieces  of  molding  made  of  wood,  miters 
at  the  same  angle  be  cut  (right  and  left)  by  means  of  a 
saw,  and  that  they  be  then  placed  together,  as  shown 
in  Fig.  239.  Now,  if  a  piece  of  sheet  iron,  for  ex- 
ample, be  slipped  into  the  joint,  as  shown  by  A,  and 
then  one  arm  of  the  miter  be  removed  what  is  left  will 
be  exactly  what  is  shown  in  Fig.  238.  In  other  words, 
a  miter  between  two  straight  pieces  of  molding  of  the 
same  profile  is  exactly  the  same  as  a  miter  of  the  same 
mold  against  a  plane,  and,  hence,  the  operation  of 
cutting  the  pattern  in  such  a  case  as  shown  in  Fig. 
239  is  identical  with  that  described  in  Figs.  237  and 
238. 

From  this  it  is  plain  to  be  seen  that  the  central 
idea  in  miter  cutting  is  to  bring  the  points  from  the 
profile  against  the  miter  line,  no  matter  what  may  be 
its  shape  or  position,  and  from  the  miter  line  into  a 
stretch-out  prepared  to  receive  them.  Inasmuch  as  all 
moldings,  if  they  do  not  member  or  miter  with  dupli- 
cates of  themselves,  must  either  terminate  square  or 
against  some  dissimilar  profile,  it  follows  that  the  two 
illustrations  given  cover  in  principle  the  entire  cata- 
logue of  miters. 

The  principles  here  explained  are  the  funda- 
mental principles  in  the  art  of  pattern  cutting,  and 
their  application  is  universal  in  sheet  metal  work.  It 


would  be  difficult  to  compile  a  complete  list  of  miter 
problems.  New  combinations  of  shapes  and  new  con- 
ditions are  continually  arising.  The  best  that  can  be 
done,  therefore,  in  a  book  of  this  character,  is  to  pre- 
sent a  selection  of  problems  calculated  to  show  the 
most  common  application  of  principles  which,  care- 
fully studied,  will  so  familiarize  the  student  with  them 
that  he  will  have  no  difficulty  afterward  in  working 
out  the  patterns  for  whatever  shapes  may  come  up  in 
his  practice,  whether  they  be  of  those  specifically  il- 
lustrated or  not. 

From  the  foregoing  the  following  summary  of 
requirements,  together  with  a  general  rule  for  cutting 
all  miters  whatsoever,  are  derived  : 

Requirements. — There  must  be  a  plan,  elevation 
or  other  view  of  the  shape,  showing  the  line  of  the 
joint  or  surface  against  which  it  miters,  in  line  with 
which  must  be  drawn  a  profile  or  sectional  view  of 
same,  and  this  profile  must  be  prepared  for  use  bv 
having  all  its  curved  portions  divided  into  such  a 


Fig.  24D.— Usual  Method  of  Cutting  a  Square  Return  Miter. 

number  of  spaces  as  is  consistent  with  accuracy  and 
convenience. 

It  may  be  remarked  here  that  the  division  of  the 
profile  into  spaces  is  only  an  approximate  method  of 
obtaining  a  stretch-out.  As  theoretically  the  straight 
distance  from  one  of  the  assumed  points  to  another 
upon  a  curved  line  is  less  than  the  distance  measured 
around  the  curve,  and  the  shorter  the  radius  of  the 


Principles  of  Pattern   Cutting. 


77 


PROFILE 


curve  the  greater  is  this  difference  (a  chord  is  le.s.s  than 
the  arc  which  it  subtends,)  hence  the  greater  the  num- 
ber of  points  assumed  the  greater  will  be  the  accuracy, 
and  a  curve  of  short  radius  should  be  divided  more 
closely  than  one  of  longer  radius.  The  profile  thus 
represents  practically  a  succession  of  plane  surfaces. 

Rule. — 1.  Place  a  stretch-out  of  the  profile  on  a 
line  at  right  angles  to  the  direction  of  the  molding,  as 
shown  in  the  plan,  elevation  or  other  view,  through 

the    points    in  which   draw 
1 t«i  L 

measuring  lines  parallel  to 
|N  the  molding.      2.  Drop  lines 

I    \ii  nor,CM   p 

from  the  points  in  the  profile 
to  the  miter  line  or  line  of 
joint,  carrying  them  in  the 
direction  of  the  molding  till 
they  intersect  said  line.  3. 
Drop  lines  from  the  inter- 
sections thus  obtained  with 
the  miter  or  joint  line  on  to 
the  measuring  lines  of  the 
stretch-out,  at  right  angles 
to  the  direction  of  the  mold- 
ing. 

In  making  the  applica- 
tion of  this  rule  the  student 
must  not  forget  that  the 
word  profile  covers  a  vast 


PA 


Fvj.  ^41.— Comparison  Between  the.  Short  or   Usual  Method  of  Cutting  a  Square 
Miter  and  the  Method  Prescribed  by  the  Rule. 


range  of  outlines,  varying  from  a  simple  straight  line 
to  an  entire  section  of  a  roof  or  even  more,  where  large 
curved  surfaces  are  to  be  treated,  and  that  a  rule  that 
applies  to  one  can  be  applied  to  the  others  equally 
well. 

The  student  who  gives  careful  attention  to  these 
rules  will  at  once 'remark  that  the  operation  of  cutting 
a  common  square  miter — that  is,  a  miter  between  the 
moldings  running  across  two  adjacent  sides  of  a  square 


building,  for  example — does  not  employ  a  miter  line, 
and,  therefore,  appears  to  be  an  exception.  Yet  it 
has  been  remarked  that  a  thorough  understanding  of 
how  a  square  miter  is  cut  comprehends  within  itself 
the  science  of  miter  cutting.  The  square  return  miter 
— for  such  is  the  distinctive  name  applied  to  the  kind 
of  square  miter  in  question — is  an  exception  to  the 
general  rule  only  in  the  sense  that  it  admits  of  an 
abbreviated  method.  The  short  rule  for  cutting  it  is 
usually  the  first  thing  a  pattern  cutter  learns,  and  the 
operation  is  very  generally  explained  to  him  without 
any  reason  being  given  for  the  several  steps  taken. 
In  many  cases  it  would  bother  him  to  cut  the  pattern 
by  any  other  than  the  short  method,  even  after  he  had 
obtained  considerable  proficiency  in  his  art.  Hence  it 
is  that,  to  all  who  have  any  previous  knowledge  of 
pattern  ciitting  the  rules  above  set  forth  seem  inade- 
quate, or,  to  put  it  otherwise,  are  a  formula  to  which. 
there  are  exceptions. 

To  clear  up  these  doubts  in  the  mind  of  the  stu- 
dent an  illustration  of  the  short  method  of  cutting  a 
square  miter  is  here  introduced,  and  afterward  the  long 
method,  or  the  plan  which  is  in  strict  accordance  with 
the  rule  above  given,  will  be  presented,  combined  with 
the  short  method,  thus  showing  the  relationship  and 
correspondence  between  the  two. 

Fig.  2iO  shows  the  usual  method  of  developing  a 
square  return  miter,  being  that  in  which  no  plan  lino 
is  employed.  The  profile  A  B  is  divided 
into  any  convenient  number  of  spaces,  as 
indicated  by  the  small  figures  in  the  en- 
graving. The  stretch-out  E  Fis  laid  off  at 
right  angles  to  the  lines  of  the  moldings, 
and,  through  the  points  in  it,  measuring 
lines  are  drawn  parallel  to  the  lines  of 
moldings.  From  the  points  established  in 
the  profile  lines  are  dropped  cutting  corre- 
sponding measuring  lines.  Then  the  pattern 
or  miter  cut  G  II  is  obtained  by  tracing  a 
line  through  these -points  of  intersection. 

In  this  operation  it  will  be  noticed  that 
the  stipulations  of  the  first  part  of  the  rule  have  been 
fully  complied  with — that  is,  the  stretch-out  line  has 
been  drawn  at  right  angles  to  the  lines  of  the  molding, 
and  measuring  lines  have  been  drawn  parallel  to  those 
lines,  but  it  would  seem  that  the  second  and  third  parts 
of  the  rule  as  given  are  not  applicable.  Apparently  no 
miter  line  has  been  employed,  but  the  points  have 
been  dropped  directly  from  the  profile  into  the  measur- 
ing lines. 


78 


Sete  M'-tai    H  "(,/•/•(/•  Pattern 


in  order  to  make  this  clear  Fig.  24-1  is  hero  in- 
troduced in  which  the  proper  relation  of  parts  is  shown 
and  in  which  the  pattern  is  developed  according  to 
rule,  and  in  which  is  also  shown  the  short  method  and 
how  it  is  derived  from  the  long  method. 

As  the  angle  of  a  return  miter  can  only  be  shown 
by  a  plan,  the  plan  becomes  the  first  necessity  accord- 
ing to  the  rule  and  is  shown  in  the  cut  by  H  F  K  M 
G  L,  F  G  showing  the  lino  upon  which  the  two  anus 
of  the  molding  meet — that  is,  the  miter  line.  The 
profile  A  B  appears  duly  in  line  with  one  arm  of  the 
plan  H  F  G  L.  This  arm,  then,  is  the  part  of  which 
the  pattern  is  about  to  be  developed  ;  accordingly  the 
stretch-out  line  is  then  drawn  at  right  angles  to  this 
arm,  as  shown  at  C'  I)',  and  the  measuring  lines  drawn 
parallel  to  the  arm. 

The  second  part  of  the  rule  is  now  carried  out; 
that  is,  lines  are  dropped  from  the  points  in  the  profile 
A  B  to  the  miter  line  F  G  and  from  thence  at  right 
angles  to  F  II  into  the  measuring  lines,  thus  obtaining 
the  pattern  C'  F/. 

In  the  upper  part  of  this  figure  another  stretch-out, 
C  D,  is  introduced  into  which  lines  have  been  dropped 
directly  from  the  points  in  the  profile,  thus  producing 
the  pattern  at  C  E,  making  this  part  of  the  figure  a 
reduplication  of  the  method  employed  in  the  previous 
figure. 

By  comparison  it  will  be  seen  that  the  two  patterns 
0  E  and  C'  E'  are  identical.  Since  the  two  arms  of 
the  miter  are  identical  and  at  right  angles  to  each  other, 
the  miter  line  must  bisect  the  angle  II  F  K  and  be  at 
an  angle  of  4-5  degrees  to  either  of  the  two  faces  II  F 
and  F  K.  From  this  it  appears  at  once  that  the  pro- 
jection of  any  and  all  points  upon  F  G  from  the  plan 
line  G  L  toward  II  is  exactly  the  same  as  from  the 
plan  line  G  M  toward  K  and  that  the  relationship  be- 
tween C  E  and  the  miter  line,  and  C'  E'  and  the  miter 
line,  is,  therefore,  the  same.  Dropping  points  from 
a  profile  against  a  line  inclined  45  degrees,  as  F  G,  and 
thence  on  to  a  stretch-out,  gives  the  same  result  as 
dropping  them  on  the  stretch-out  in  the  first  place. 
Hence  it  is  that  the  portion  of  the  operation  shown  in 
the  lower  part  of  the  engraving  may  be  dispensed  with. 
This  relationship  could  never  occur  were  the  angle  of 
the  miter  anything  else  than  a  right  angle. 

Another  and  perhaps  simpler  explanation  of  this 
is  given  in  connection  with  Problem  3,  in  Section  1  of 
Chapter  VI. 

A  very  common  mistake  made  by  beginners  in 
attempting  to  apply  the  general  rule  for  cutting  miters 


as  given,  is  that  of  getting  the  miter  line  in  a  wrong 
position  with  reference  to  the  profile.  For  example, 
instead  of  drawing  a  complete  plan,  as  shown  bv  L  II 
F  K  M  in  Fig.  i>41,  by  which  the  miter  line  is  located 
to  a  certainty,  and  in  connection  with  which  it  is  a 
simple  matter  to  correctly  place  the  profile,  it  is  not 
uncommon  to  attempt  the  operation  by  drawing  the 
miter  line  only,  placing  it  either  above,  below  or  at 
one  side  of  the  profile.  The  mistake  is  made  bv  hav- 
ing the  line  at  the  side  of  the  profile  when  it  should 
be  either  above  or  below  it,  and  vice  versa.  Fig.  -24-2 
illustrates  a  ease  in  point.  The  engraving  was  made 


L  .  B    , 


Fig.  24%- — A.  Squure  Face  Miter  Produced  Where   a   Square  Return 
Miter  ivas  Intended, 


from  the  drawing  of  a  person  who  attempted  to  cut  a 
square  return  miter  by  the  rule,  using  a  miter  line 
only.  By  placing  the  miter  line  E  F  at  the  side  in- 
stead of  below  the  profile,  a  square  face  miter — such 
as  would  be  used  in  the  molding  running  around  a 
panel  or  a  picture  frame — was  produced  in  place  of 
what  was  desired. 

In  order  to  avoid  such  errors  the  reader  is  recom- 
mended to  a  careful  pei'usal  of  the  chapter  on  Linear 
Drawing  (Chapter  III),  where  the  relation  existing  be- 
tween plans,  elevations  and  sections  or  profiles  is  thor- 
ough- explained.  It  is  better  to  draw  a  complete  plan. 
as  :-hown  in  Fig.  24-1,  thus  demonstrating  to  a  eer- 


Principles  uf  Pattern    Cuttinij. 


tainty  the   correct  relationship  of    the  parts,  than    to 
save  a  little  labor  and  run  the  risk  of  error. 

As  remarked  in  the  earlier  part  of  this  chapter, 
some  labor  is  often  neeessarv  before  the  requirements 
mentioned  above  in  connection  with  the  rule  can  be 
fulfilled.  Sometimes  a  miter  line  must  first  be  de- 
veloped, and  sometimes  the  profile  of  a  molding  must 
undergo  a  change  of  profile  known  as  raking.  It  is 
believed  that  the  principles  underlying  these  opera- 
tions arc  made  sufficiently  dear  in  connection  with  the 
problems  in  which  they  are  involved  not  to  need  es- 
pecial explanation  in  this  connection.  Suffice  it  to 
say  that,  in  many  instances,  half  the  work  is  done  in 
the  getting  ready. 


(FLARING    WORK.) 

This  subject  embraces"  a  large  variety  of  forms  of 
frequent  occurrence  in  sheet  metal  work,  and  the  de- 
velopment of  their  surfaces  comes  under  an  altogether 
different  set  of  rules  than  those  applied  to  parallel 
forms. 

Before  entering  into  the  details  of  these  methods 
it  will  be  best  to  first  define  accurately  what  is  here 
included  by  the  use  of  the  term.  These  forms  include 
only  such  solid  figures  as  have  for  a  base  the  circle  or 
any  of  the  regular  polygons,  as  the  square,  triangle, 


Fig. 


.  —  A  Right  Cone   Generated  by  the  Revolution   of  a  Right- 
Angled  Triangle  about  its  Perpendicular. 


hexagon,  etc.  ;  also  figures  though  of  unequal  sides 
that  can  be  inscribed  within  a  circle,  and  all  of  which 
terminate  in  an  apex  located  directly  over  the  center 
of  the  base. 

While  the  treatment  of  these  forms  has  been  said 
to  be  altogether  different  from  that  of  parallel  forms 
there  are  some  points  of  similarity  to  which  the  stu- 
dent's attention  is  called  that  may  serve  to  fix  the 
methods  of  work  in  his  memory. 


Whereas  in  parallel  forms  the  -distances  of  the 
various  [mints  in  a  miter  are  measured  from  a  straight 
line  drawn  through  the  mold  near  the  miter  for  that 
purpose,  as  C  D,  Fig.  23.8,  the  distances  of  all 
points  in  the  surfaces  of  tapering  solids  produced  by 
the  intersection  of  some  other  surface  are  measured 
from  the  apex  upon  lines  radiating  therefrom ;  and 


Fig.  244.— A  Bight  Cone  with  Thread  Fastened  at  the  Apex  to  which 
are  Attached  Points  Marking  the  Upper  and  Lower  Bases  of  a 
Frustum. 


whereas  the  distance  across  parallel  forms  (the  stretch- 
out) is  measured  upon  the  profile,  the  distance  across 
tapering  forms  is  measured  upon  the  perimeter  of  the 
base. 

Patterns  are  more  frequently  required  for  por- 
tions of  frustums  of  these  figures  than  for  the  com- 
plete figures  themselves  and  the  methods  of  obtaining 
the  pattern  of  coverings  of  said  frustums  is  simply  to 
develop  the  surface  of  the  entire  cone  or  pyramid  and 
by  a  system  of  measurements  take  out  such  parts  as 
are  required. 

As  the  apex  of  a  cone  is  situated  in  a  perpendic- 
ular line  erected  upon  the  center  of  its  base,  it  must 
of  necessity  be  equidistant  from  all  points  in  the  cir- 
cumference of  the  base. 

In  works  upon  solid  geometry  the  cone  is  de- 
scribed as  a  solid  generated  by  the  revolution  of  a 
right-angle  triangle  about  >ts  vertical  side  as  an  axis. 
This  operation  is  illustrated  in  Fig.  243,  in  which  it 
will  be  seen  that  the  base  E  1)  of  the  triangle  C  K  1) 
is  the  radius  which  generates  the  circle  forming  the 
base  of  the  cone,  and  that  the  hypothenuse  C  D  in 
like  manner  generates  its  covering  or  envelope. 


80 


The  Xcw  Metal    Wurkvr   I'aW_/-n  Book. 


If  a  plane  bo  passed  through  u  00110  parallel  to 
the  base  and  at  some  distance  above  it,  the  line  which 
it  produces  by  cutting  the  surface  of  the  cone  must 
also  be  a  circle,  because  it,  like  the  base,  is  perpen- 
dicular to  the  axis.  The  portion  cut  away  is  simply 
another  perfect  cone  of  less  dimensions  than  the  first. 


Fig.  S45. — Frustum  of  a  Right  Cone,  the  Dotted  Lines  Showing  the 
Portion  of  the  Cone  Removed  to  Produce  the  Frustum. 

while  the  portion  remaining  is  called  a  frustum  of  a 
cone.  AFC,  Fig.  244,  is  a  cone,  and  B  D  E  C,  Fig. 
245,  is  a  frustum.  The  line  B  E,  Fig.  244,  shows 
where  the  cone  is  cut  to  produce  the  frustum. 

If,  having  a  solid  cone  of  any  convenient  material, 
as  wood,  a  pin  be  fastened  at  tlie  apex  C  of  the  same, 
as  shown  in  Fig.  244,  and  a  piece  of  thread  be  tied 


Fig.  246. — Envelope  of  the  Cone  and  Frustum,  Described  by  the  Pin 
and  Thread  in  Fig.  244. 

thereto,  to  which  are  fastened  points  B  and  A,  corre- 
sponding in  distance  from  the  apex  to  the  upper  and 
lower  bases  of  the  frustum,  and  the  thread,  being  drawn 
straight,  be  passed  around  the  cone  close  to  its  surface, 
the  points  upon  the  thread  will  follow  the  lines  of  the 
liases  of  the  frustum  throughout  its  course.  If  then. 
taking  the  thread  and  pin  from  the  cone,  and  fastening 


the  pin  as  a  center  upon  a  sheet  of  paper,  as  shown  in 
Fig.  24(5,  the  thread  be  carried  around  the  pin,  keep- 
ing it  stretched  all  the  time,  the  track  of  the  points 
fastened  to  the  thread  will  describe  upon  the  paper  the 
shape  of  the  envelope  of  the  frustum,  as  shown  by 
(J  D  E  F.  By  omitting  the  line  produced  bv  the 
upper  of  the  two  points,  the  envelope  of  the  complete 
cone  G  C  F  will  be  described.  The  length  of  the  arc 
<i  !•'  described  by  the  point  A  attached  to  the  thread 
may  be  determined  by  measuring  the  circumference  of 
the  base  of  the  cone  by  any  means  most  available. 
The  usual  method  is  to  take  between  the  points  of  the 
dividers  a  small  space  and  step  around  the  circumfer- 
ence of  the  circle  of  the  base  and  set  olf  upon  the  circle 
of  the  pattern  the  same  number  of  spaces. 


Fig.  2y/. — Unfolding  the  Envelope  of  a  Right  Cone. 

The  development  of  the  envelope  of  a  cone  may 
be  further  illustrated  by  supposing  that,  in  the  case 
of  the  wooden  model,  it  be  laid  upon  its  side  upon  a 
sheet  of  paper  and  rolled  along  until  it  has  made  one 
complete  revolution;  a  point  having  been  previously 
marked  upon  the  line  of  its  base  by  which  to  deter- 
mine the  same.  The  base  B,  Fig.  247,  thus  becomes 
stretched  out  as  it  were,  describing  the  line  C  D  upon 
the  paper,  while  the  apex  A,  having  no  cireumfer- 
ference,  remains  stationary  at  the  point  A'.  The  lines 
C  A'  and  D  A1  represent  the  contact  of  the  side  of  the 
cone  at  the  beginning  and  at  the  finish  of  one  revolu 
tion. 

As  in  the  case  of  dividing  the  prolilc  in  parallel 
forms, -this  method  is,  theoretically,  only  approximate 
in  accuracy,  but  the  difference  is  so  slight  practically 
that  it  is  not  worth  considering.  Of  course,  the  shorter 


Principles  <</"  Pattern   O/////y. 


81 


the  spaces  are  the  greater  is  the  accuracy.  This 
method  lias,  however,  another  significance  which  will 
bo  pointed  out  later  on,  which  will  help  to  simplify 
the  solution  of  all  tapering  forms. 


Fig.  24$.— A  Cone  Truncated  Obliquely. 

If  it  is  required  that  the  cone  should  be  truncated 
obliquely,  as  shown  in  Fig.  248,  it  will  be  seen  that 
all  the  points  in  the  upper  line  of  the  .frustum  are  at 
different  distances  from  the  base,  or,  what  amounts  to 
the  same  thing,  from  the  apex  of  the  original  cone, 
hence  some  method  of  measuring  these  distances  must 
be  devised. 

To    explain    the    principles    here   involved   more 


H 


-r if 


Fig.  249. — Plan  and   Elevation   from  which   to  Construct  a  Wooden 
Cone  for  Purposes  of  Illustration. 

clearly,  suppose  that  a  cone  be  cut  from  a  solid  block  of 
wood  and  of  a  hight  and  width  to  agree  with  some  par- 
ticular drawing,  as,  for  instance,  the  one  shown  in  Fig. 
24-9.  Divide  the  circle  of  the  base  E  F  upon  the  drawing 
into  a  convenient  number  of  parts  or  spaces  and  mark 


the  same  number  of  points  and  spaces  upon  the  edge 
of  the  base  of  the  wooden  cone,  and  from  each  of  these 
points  draw  upon  the  sides  of  the  wooden  cone  straight 
lines  running  to  its  apex. 

A  correct  elevation  of  these  lines  upon  the  draw- 
ing may  be  obtained  by  carrying  lines  from  the  divisions 
or  points  in  the  plan  of  the  base  vertically  till  they 
strike  the  line  of  the  base  B  C  in  the  elevation,  as  shown 
in  Fig.  250,  thence  to  the  apex  A,  cutting  the  line 
G  H. 

Now,  if  by  means  of  a  saw  the  upper  part  of  the 
wooden  cone  be  removed,  being  cut  to  the  required 
angle  as  shown  by  the  oblique  line  G  H  in  the  draw- 
ing, an  opportunity  is  given,  by  the  lines  upon  the 


Fig.  250.— Method  of  Obtaining  the  Lines  upon  the  Elevation. 

part  of  the  cone  cut  away,  of  measuring  accurately  the 
distance  of  each  point  of  the  curve  thus  produced  from 
the  apex. 

Then  as  all  points  in  the  base  B  C  are  equidistant 
from  the  apex  A,  to  lay  out  the  pattern  of  this  frustum, 
first  describe  an  arc  of  a  circle  whose  radius  is  equal  to 
the  length  of  the  side  (or  slant  hight)  of  the  cone  A  B, 
Fig.  250.  Make  this  arc  in  length  equal  to  the  cir- 
cumference of  the  base  B  C  of  the  cone  by  means  of 
the  points,  as  previously  described.  To  avoid  con- 
fusion number  these  points  1,  2,  3,  etc.,  from  the  start- 
ing point  B,  and  from  each  of  these  points  draw  lines 
to  the  center  of  the  arc,  all  as  shown  in  Fig.  251. 

Now,  replacing  that  portion  of  the  cone  which 
was  cut  away  so  as  to  identify  the  lines  upon  its  siites 


82 


T/ie  Xeic  Metal    Worker  Pattern   Book. 


by  the  numbers  at  the  base,  the  length  of  each  line 
from  the  apex  down  to  the  cut  can  be  measured  by  the 
dividers  and  transferred  to  the  lines  of  the  same  num- 
bers in  the  diagram,  Fig.  251,  as  shown  between  G 
and  H. 

All  this  no  doubt  is  quite  simple  when  the  model 
is  at  hand  upon  which  to  make  the  measurements.  It 
;  is  quite  evident  that  it  will  not  do  to  measure  the  dis- 
tance upon  the  drawing,  Fig.  250,  from  the  apex  A  to 
the  points  of  intersection  on  the  line  G  H  because  the 
sides  of  the  cone  having  an  equal  slant  of  flare  all 
around,  the  lines  upon  the  drawing  do  not  represent 
the  real  distances  except  in  the  case  of  the  two  outside 
lines;  the  slant  hight  of  a  cone  or  any  part  of  a  cone 
being  greater  than  the  vertical  hight  of  same  part. 
But  as  these  two  outside  lines  do  represent  the  correct 


G 


B        9 


Fig.  Sol. — Method  of  Deriving  the  Pattern  of  a  Frustum  from  the 
Wooden  Model. 

slant  of  the  cone  on  all  sides,  either  one  of  them  may 
be  taken  as  a  correct  line  upon  which  to  measure  these 
distances ;  that  is,  as  a  vertical  section  through  the 
cone  upon  any  or  all  of  the  lines  drawn  upon  its  sides. 
To  make  it  a  perfect  section  upon  any  one  of 
these  lines,  say  line  5,  it  is  simply  required  that  the 
position  of  the  point  of  intersection  of  line  5  with  the 
line  G  H  be  shown,  which  is  done  by  carrying  this 
point  horizontally  across  till  it  strikes  the  side  of  the 
cone  A  B  at  5,  as  illustrated  in  Fig.  250.  The  result 
of  repeating  this  operation  upon  all  the  other  lines  is 
as  though  a  thread  or  wire  were  stretched  from  the 
apex  down  along  the  side  of  the  cone  to  the  point  B 
in  the  base  and  the  cone  were  turned  upon  its  axis, 
and  as  each  line  upon  the  side  passes  under  the  thread, 
the  point  where  it  cuts  the  intersecting  plane  <!  II 
where  marked  thereon,  thus  collecting  all  the  points 
inft)  one  section  as  it  were. 


This  operation  is  fully  shown  in  Fig.  252,  to 
which  is  added  the  development  of  the  pattern,  which 
is  exactly  the  same  as  that  shown  in  Fig.  251,  the  dis- 
tances of  the  points  between  G  and  II  from  A  being 
obtained  in  this  case  from  the  points  upon  the  line 
A  B,  instead  of  from  the  model,  as  before.  The  points 
on  A  B  are  transferred  to  lines  of  corresponding  num- 
ber in  the  pattern  by  means  of  the  compasses,  as 
shown. 

Should  the  frustum  of  which  a  pattern  is  required 


Fig.  %52. — Method  of  Deriving  the  Pattern  from  >/ic  Dm  winy. 

have  both  its  upper  and  lower  faces  oblique  to  the 
axis  of  the  cone  a  level  base  can  be  assumed  at  a  con- 
venient distance  below  the  lower  face  of  the  frus- 
tum from  which  the  circumference  can  lie  obtained 
and  then  both  the  upper  and  lower  faces  of  the  frus- 
tum can  be  developed  by  the  method  just  described. 

A  right  cone  having  an  elliptical  base  might  seem 
to  belong  in  the  same  class  with  regular  tapering  forms, 
but  as  the  distance  from  its  apex  to  the  various  points 
in  the  perimeter  of  its  base  is  constantly  varying,  it  is 
therefore  placed  in  the  class  with  irregular  forms  in 


Principles  <y  Pattern 


83 


tin-  following  section  of  this  chapter,  where  it  will  be 
duly  diseiisseil.  But  as  tapering  articles  of  elliptical 
shape  are  of  frequent  occurrence,  and  as  the  circle  is 
much  easier  made  use  of  than  the  ellipse,  such  articles 
are  usually  designed  with  approximate  ellipses  com- 
posed ul  arc-sol  circles.  This  method  is  in  inanv  cases 
especially  desiralile,  as  articles  so  designed  have  an 
equal  amount  ol  Hare,  or  taper  on  all  sides,  which 
would  not  he  the  case  if  they  were  cut  from  elliptical 
cones.  It  will  thns  he  seen  that  an  article,-  designed  in 
I  hat  manner  is  the  envelope  of  a  solid  composed  of  as 
many  portions  of  frustums  of  right  cones  as  there  were 
arcs  of  circles  used  in  drawing  its  plan. 

In  Fig.  2.~>:>  is  shown  the  usual  method  of  draw- 
ing the  plan  and  elevation  of  an  elliptical  flaring  ar- 
ticle, the  outer  curve  of  the  plan  A  C  B  D  being  the 
shape  at  M  N  of  the  elevation,  while  the  inner  curve 
I  G  A'  J  is  the  plan  at  the  top  K  L.  As  many 
centers  may  he  employed  in  drawing  the  curves  of  the 
plan  of  such  an  article  as  desired,  all  of  which  is  ex- 
plained in  the  chapter  on  Geometrical  Problems  (Chap. 
I V. ),  Problems  73,  70  and  78.  To  simplify  matters  only 
two  sets  of  centers  have  been  employed  in  the  present 
drawing,  all  as  indicated  by  the  dotted  lines  drawn  from 
the  various  centers  and  separating  the  different  arcs  of 
circles.  Keference  to  the  plan  now  shows  that  that 
portion  of  the  article  included  between  the  points  E 


Fi'j.  253.— Usual  Method  of  Drawing  an  Elliptical  Flaring  Article. 

C  W  V  G  II  is  a  position  of  the  envelope  of  a  cone 
the  radius  of  whose  base  is  1)  C  and  whose  apex  is 
located  at  a  point  somewhere  above  D ;  and  likewise 
that  that  portion  included  between  the  points  X  A  B 
II  I  J  is  part  of  a  cone  the  radius  of  whose  base  is 
A  F  and  whose  apex  is  somewhere  upon  a  line  erected 
at  F.  Thus  four  sectors  cut  from  cones  of  two  differ- 
ent sizes  go  to  make  up  the  entire  solid  of  which  the 
Article  shown  in  Fig.  253  is  a  frustum. 


To  determine  the  dimensions,  then,  of  such  cones 
it  is  necessary  to  construct  a  diagram  such  as  that 
shown  in  Fig.  254,  which  is  in  reality  a  section  upon 
the  line  E  D  of  the  plan,  in  which  P  0  and  P  E 
are  respectively  equal  to  E  D  and  E  F  of  the  plan. 
At  points  0  and  R,  Fig.  254,  erect  perpendicular  lines 
0  J  and  K  Z  indefinitely.  Upon  O  J  set  off  OS  equal 
to  the  straight  hight  O  K  of  the  frustum,  Fig.  253,  and 


Fiy.  2J4.—Diayram  Constructed  to  Determine  the  Dimensions  of  the 
Cones,  Portions  of  which  are  Combined  to  Make  up  the  Article 
Shown  in  Fig.  253. 

draw  S  U  parallel  to  O  P,  which  make  equal  in  length 
to  D  H.  A  line  drawn  through  the  points  P  and  U 
will  then  represent  the  slant  or  taper  of  the  frustum, 
as  shown  at  M  K  of  the  elevation,  and  if  continued  till 
it  intersects  with  the  perpendiculars  from  O  and  P  will 
determine  the  respective  hights  of  the  two  cones,  as 
shown  by  Z  and  J.  Then  P  JO  is  the  triangle  which, 
if  revolved  about  its  vertical  side  J  0,  will  generate 
the  cone  from  which  so  much  of  the  figure  as  is  struck 
from  the  centers  C  and  D  in  Fig.  253  is  cut ;  and  P  Z  R 
is  the  triangle  which  if  revolved  about  its  vertical  side 
Z  R  will  generate  the  cone  from  which  the  end  pieces 
of  the  article  are  taken.  To  present  this  before  the 
reader  in  a  more  forcible  manner,  several  pictorial  illus- 
trations are  here  introduced  in  which  the  foregoing 
operations  are  more  clearly  shown.  In  Fig.  255  is 
shown  a  view  of  the  plan  of  the  base  A  C  B  D  of  Fig. 
253  in  perspective,  in  which  the  reference  letters  are 
the  same  as  at  corresponding  parts  of  that  plan,  and 
upon  which  is  represented,  in  its  correct  position,  a 
sector  of  the  larger  cone  from  which  the  side  portions 


The  Xvw  Mini    \Vnrkcr  Pattern  Buok. 


of  the  frustum  are  taken.  Tims  the  triangular  sur- 
faces. F  D  E  and  F  D  AV,  being  sections  of  the  cone 
through  its  axis,  correspond  to  the  triangle  J  ()  P  of 
the  diagram,  Fig.  254.  In  Fig.  256  two  additional 
sectors  from  the  smaller  cone  previously  referred  to 
are  represented  as  standing  upon  the  adjacent  portions 
of  the  plan  from  which  their  dimensions  were  derived. 
Thus  C  F  and  II  D,  the  center  lines  of  their  bases, 
correspond  respectively  to  A  F  and  Y  B  of  the  plan, 
Fig.  253,  and  the  triangles  L  F  G  and  M  H  K,  being- 
radial  sections  of  the  cones,  correspond  with  the  tri- 
angle Z  R  P  of  the  diagram.  In  Fig.  257  is  presented 
the  opposite  view  of  the  combination  seen  in  Fig.  250, 


Fig.  255. — Perspective  View  of  the  Plan  in  Fig.  253,  with  a  Sector  of 
the  Larger  Cone  in  Position, 

showing  at  C  B  and  D  A  the  joining  of  their  outer 
surfaces  or  envelopes. 

As  previously  remarked,  two  sets  of  centers  only 
were  employed  in  constructing  the  plan,  Fig.  253,  for 
the  sake  of  simplicity.  Had  a  third  set  of  centers 
been  made  use  of  the  arrangement  of  sectors  of  cones 
shown  in  Figs.  256  and  257  would  have  been  supple- 
mented by  a  pair  of  sectors,  cut  from  a  cone  of  inter- 
mediate size,  which  would  have  been  placed  on  either 
side  of  the  large  sector  between  it  and  the  smaller 
ones,  all  being  joined  together  upon  the  same  general 
principle  as  before  explained.  Reference  to  Fig.  22<) 
in  the  chapter  on  Geometrical  Problems  shows  at  .1  L 
P,  P  S  W  and  AV  IT  Y  what  the  relative  position  of 
their  bases  would  be.  If  it  be  desired  to  complete  the 
solid,  which  has  been  begun  in  Fig.  256,  it  will  first  be 
.necessary  to  cut  away  that  portion  of  the  middle  sector 


which  stands  over  the  space  F  H  B,  Fig.  256.  Such  a 
cut  might  be  begun  upon  the  line  F  H,  and  passing 
vertically  through  the  points  L  and  M  would  finish 
through  the  curved  surface  of  the  further  or  curved 
side  of  the  sector.  The  cut  thus  made  between  the 
points  L  and  M  is  shown  at  D  C  in  the  other  view, 
Fig.  257,  and  is  by  virtue  of  the  conditions  a  hyper 
bola.  (See  Def.  113,  Chap.  I.)  The  piece  necessary 
to  complete  the  solid  would  then  be  a  duplicate  of  the 
shape  remaining  after  making  the  above  described  cut, 
the  outer  surface  of  which  is  shown  by  A  B  C  D  of 
Fig.  257.  The  complete  solid  would  then  have  the 
appearance  shown  in  Fig.  258. 


Fig.  256. — The  Same  Plan  Showimj  Two  Sectors  of  the  Smaller  Cone 
in  Position  Joining  the  Larger  One. 

By  thus  resolving  the  solid  from  which  the  ordi- 
nary elliptical  flaring  article  is  cut  into  its  component 
elements  the  process  of  developing  its  pattern  may  be 
more  readily  understood.  This  process  may  now  lie 
easily  explained  by  returning  to  the  string  and  pin 
method  which  was  made  use  of  in  connection  with  the 
simple  cone  in  the  earlier  part  of  this  section. 

In  Fig.  257  is  shown  a  line  some  distance  above 
the  base  representing  the  top  of  the  frustum  shown  by 
K  L  in  the  original  elevation,  Fig.  253.  It  also 
shows  a  pin  fastened  at  the  apex  of  the  middle  conical 
sector  to  which  is  attached  a  thread  carrying  points 
G  and  II  representing  the  upper  and  lower  surfaces  of 
the  frustum.  Now,  if  the  siring  be  drawn  tight  and 
passed  along  the  side  of  the  larger  sector  of  the  cone 
from  A  to  B  the  points  will  follow  the  upper  and 
lower  bases  of  the  frustum.  AVhen  the  point  B  is 


Principles  «/'  /'"//</'/'    Cull! HI/. 


85 


reached,  if  the  finger  be  placed  upon  the  thread  at  the 
apex  of  the  lesser  emie,  shown  at  C,  and  the  progress 
of  the  thread  be  eontinued,  the  points  will  still  follow 
the  lines  of  the  bases,  of  the  frustum.  If  the  pin  and 


Fly.  S»7. — Opposite  View   of  Parts   Shown    in    Fig.  256,  with    String 
Attached  to  a  Pin  (it  the  Apex  of  the  Larger  Sector. 

thread  be  taken  from  the  cone  and  transferred  to  a 
sheet  of  paper,  as  shown  in  Fig.  250,  the  pin  A  being 
used  as  a  center  and  the  thread  as  a  radius,  the  points 
will  describe  the  envelope  of  the  frustum.  First,  the 
radius  is  used  full  length,  as  shown  by  A  L  K,  and 
arcs  L  M  and  K  H  are  drawn  in  lengtlTrespectively 


Fii/.   258.— The    Completed  Solid   of  which   the   Ordinary   Elliptical 
Flaring  Article  in  a  Pnrt. 

equal  to  their  representatives  II  (T  V  and  E  C  \\  of 
the  original  plan,  Fig.  253.  Then  a  second  pin  is 
put  through  the  string,  as  shown  at  B,  thus  reducing 
the  radius  to  the  length  of  the  side  of  the  lesser  cone,^ 
and  ares  are  struck  in  continuation  of  those  first  de- 


scribed, making  the  length  of  the  additional  arcs 
equal  to  those  of  their  corresponding  arcs  H  I  J  and 
K  A  X  of  the  original  plan. 

As   the  lengths  of  the   sides  of   the   larger   and 


Fig.  259. — The  Pin  and  Thread  taken  from  Fig.  257   and    Used  in 
Describing  the  Envelope. 

smaller  cones  above  made  use  of  are  by  construction 
equal  to  J  P  and  Z  P,  the  hypothenuses  of  the  tri- 
angles, Fig.  254,  by  whose  revolution  they  were 
generated,  those  distances  may  therefore  be  taken  at 
once  from  that  diagram  by  means  of  the  compasses 
and  used  as  shown  in  Fig.  259. 

Reference  has  been  made  above  to  the  difference 
between  the  circumference  of  the  circle  of  the  base 
obtained  by  means  of  the  points  and  spaces  (which 
method  becomes  a  necessity  to  the  pattern  cutter)  and 


Fig.  260. — An  Arc  Compared  with  its  Chord. 

the  real  circumference.  An  explanation  of  this  differ- 
ence wilt  lead  to  the  next  class  of  regular  tapering 
figures — viz.  :  pyramids. 

In  the  accompanying  diagram,  Fig.  2t'.(),  ABC 
represents  the  arc  of  a  circle  of  which  the  straight  line 
A  C  is  the  chord,  being  the  shortest  distance  between 
the  two  points  A  and  C.  Therefore,  when  dividing 
a  circle  by  means  of  points  for  purposes  of  measure- 
ment, the  pattern  cutter  is  in  reality  using  a  number 
of  chords  instead  of  the  arcs  which  they  subtend. 


Tin'   New    Metal    1 1  War  Pattern    Boole. 


In  the  practice  of  obtaining  the  circumference  or 
stretch-out  of  a  circle  the  space  assumed  as  the  unit  of 
measure  should  be  so  small  that  there  is  no  perceptible 
curve  between  the  points  and,  of  course,  no  practical 
difference  between  the  length  of  the  chord  and  the 
length  of  the  arc. 

It  will  thus  be  seen  that  the  circle  representing 
the  base  of  a  cone  has  in  reality  become  in  the  hands 
of  the  pattern  cutter  a  many  sided  polygon  and  that 
the  cone  is  to  him  a  many  sided  pyramid.  As  one  of 
the  conditions  in  describing  a  regular  polygon  is  that 
its  angles  shall  all  lie  in  the  same  circle,  so  the  angles 
or  hips  of  a  pyramid  must  lie  in  the  surface  of  the  cone 
whose  base  circumscribes  the  base  of  the  pyramid  and 
whose  apex  coincides  with  the  apex  of  the  pyramid. 
Viewed  in  this  light  then,  the  lines  which  were  drawn 
upon  the  outside  of  the  wooden  cone  for  the  purpose 
of  measurement  in  the  illustration  used  above  become 
the  angles  or  hips  of  a  pyramid  and  may  be  used  for 
that  purpose  in  exactly  the  same  manner. 

In  developing  the  pattern  of  a  frustum  of  a  cone 
the  line  connecting  the  points  between  G  and  II,  Fig. 
251,  is  supposed,  of  course,  to  be  a  curved  line,  while 
in  the  case  of  a  pyramid  (the  points  or  angles  of  the 
pyramid  being  further  apart  and  the  sides  of  a  pyramid 
being  flat  instead  of  curved)  the  lines  of  the  pattern 
connecting  the  points  would  be  straight  from  point  to 
point. 

Irregular    Forms. 

(TRIANGULATION.) 

In  some  classes  of  sheet  metal  work  certain  forms 
arise  for  which  patterns  are  required,  but  which  cannot 
be  classified  under  either  of  the  two  previous  sub- 
divisions. Their  surfaces  do  not  seem  to  be  generated 
by  any  regular  method.  They  are  so  formed  that 
although  perfectly  straight  lines  can  be  drawn  upon 
them  (that  is,  lines  running  parallel  with  the  form), 
such  straight  lines  when  drawn  would  not  be  parallel 
with  each  other ;  neither  would  they  slant  toward  each 
other  with  any  degree  of  regularity. 

While  in  the  systems  described  in  the  two  previous 
portions  of  this  chapter  distances  between  lines  fanning 
with  the  form  measured  at  one  end  of  an  article  govern 
those  at  the  other  end,  in  the  forms  considered  in  this 
department  these  distances  are  continually  varying  and 
bear  no  such  relation  to  each  other.  Thus  in  parallel 
forms  (moldings)  the  distance  between  any  two  lines 
running  with  the  form  is  the  same  at  both  ends  of  the 


article,  while  in  conical  shapes  all  lines  running  with 
the  form  tend  toward  a  common  center  or  vertex,  s<> 
that  the  distances  between  such  lines  at  one  end  of  the 
article'  (provided  it  does  not  reach  to  the  vertex)  bear 
a  regular  proportion  to  the  distances  between  them  at 
the  other  end.  Hence,  in  the  development  of  the 
pattern  of  an  irregular  form  it  becomes  necessary  to 
drop  all  previously  described  systems  and  simply  pro- 
ceed to  measure  up  its  surfaces,  portion  by  portion, 
adding  one  portion  to  another  till  the  entire  surface 
has  been  covered. 

To  accomplish  this  end  one  of  the  most  simple 
of  all  geometrical  problems  is  made  use  of,  to  which 
the  reader  is  referred  (Chap.  IV.,  Problem  36) — viz.  : 
To  construct  a  triangle,  the  lengtlts  of  the  three  sides  //</'//</ 
given. 

As  from  any  three  given  dimensions  only  one 
triangle  can  be  constructed,  this  furnishes  a  correct 
means  of  measurement;  and  the  solution  of  this  prob- 
lem in  connection  with  a  regular  order  and  method  of 
obtaining  the  lengths  of  the  sides  of  the  necessary  tri- 
angles constitutes  the  entire  system.  To  carry  out 
this  system  it  simply  becomes  necessary  to  divide  the 
surface  of  any  irregular  object  into  triangles,  ascertain 
the  lengths  of  their  sides  from  the  drawing,  and  re- 
produce them  in  regular  order  in  the  pattern,  and  hence 
the  term  TRIANGULATIOX  is  most  fittingly  applied  to 
this  method  of  development  of  surfaces. 

In  all  articles  whose  sides  lie  in  a  vertical  plane, 
distances  can  be  measured  in  any  direction  across  their 
sides  upon  an  elevation  of  the  article,  but  when  the 
sides  become  rounded  and  slanting  the  length  of  a 
line  running  parallel  with  the  form  cannot  be  measured 
either  upon  the  elevation  or  the  plan.  The  elevation 
gives  the  distance  from  one  end  of  the  line  to  the  other 
vertically  or  as  it  appears  to  slant  to  the  right  or  left, 
but  the  distance  of  one  end  of  the  lino  forward  or  back 
of  the  other  can  only  be  obtained  from  the  plan  which 
while  supplying  this  dimension  does  not  give  the  hight. 
Consequently  the  true  length  of  any  straight  line  lying 
in  the  surface  of  any  irregular  form  can  only  be  ascer- 
tained by  the  construction  of  a  right-angle  triangle 
whose  base  is  equal  to  the  horizontal  distance  between 
the  required  points,  and  whose  altitude  is  equal  to  the 
vertical  distance  of  one  point  above  the  other,  the 
hypothenuse  giving  the  true  distance  between  the 
points,  or,  in  other  words,  the  required  length  of  the 
line. 

For  illustration,  Fig.  261  shows  an  article  which 
|  may  be  called  a  transition  piece,  the  base  of  which, 


l'ril<l-!l>ll:-i      I!/'     I'l/llri-ll       Cl/tlilll/. 


87 


A  15  ('  D  of  the  plan,  is  a  perfect  circle  lying  in  a 
horizontal  plane.  K  II  «(  t  lie  eli!\  ation.  Its  upper  sur- 
face, however.  N  <>  I'  <^  of  the  plan,  is  elliptical  in 
shape  and  besides  being  placed  at  one  side  of  the 
center  is  also  in  an  inclined  position,  as  shown  by  I'1  <1 
of  tlin  elevation.  T<>  the  right  of  this  plan  is  another 
drawing  of  tin-  same,  A'  IV  C'  I)',  turned  one-quarter 
around  from  which,  and  the  elevation,  is  projected 
another  view,  J'  K'  L  M,  whieh  may  be  called  the 


perimeter  at  the  top.  As  F  (!.  the  distance  across  the 
top,  is  greater  than  X  P  (its  apparent  width  in  the 
plan),  the  curve  X  ( )  P  Q  evidently  does  not  give  the 
correct  distance  around  the  top,  and  therefore  a  correct 

view    of    the   top   must    1 btained.      The  method  of 

accomplishing  this  docs  not  dill'er  from  manv  similar 
operations  described  in  connection  with  parallel  forms 
and  is  clearly  shown  in  the  drawing.  Considering 
N  <)  I*  <i  as  a  correct  plan  or  horizontal  projection  of 


,ft 

.•i  ! 


fin" 
/$ 

ML 

/  /  /  /   1 1  t , 

1  2  :i    4     5  »for 

10  8 

DIAGRAM  OF 
DOTTED  LINES 


B 


Fig.  382.  Fi(r.  261. 

Plans,  Elevations,  etc.,  of  an  Irregular  Shaped  Article,  Ittustratinij  the  Principles  of  Triangulati on . 


front  and  which  will  assist  in  obtaining  a  more  perfect 
conception  of  the  shape  of  the  article.  A  comparison 
of  the  three  views  shows  that  the  slant  of  the  sides  is 
different  at  every  point,  and  that  the  only  dimensions 
of  the  article  which  can  be  measured  directly  upon  the 
drawing  arc  the  circumference  of  the  base  and  the 
slant  hight,  as  given  at  E  F,  H  G  and  L  M. 

Before  a  pattern  of  its  side  can  be  developed  it 
will  be  necessary  to  ascertain  its  width  (or  distance 
from  base  to  top)  at  frequent  intervals  and  also  its 


the  top,  one-quarter  of  it,  as  O  N,  may  be  divided  by 
any  convenient  number  of  points  and  their  distances 
from  N  P  set  off  upon  the  parallel  lines  drawn  from 
N"  P",  thus  obtaining  O"  N",  one-quarter  of  the  cor- 
rect curve.  It  is  more  likely,  however,  that  the  cor- 
rect shape  of  the  top  N""  O"  P"  would  be  given,  from 
which  it  would  be  necessary  to  obtain  its  correct 
appearance  in  the  plan,  which  would  be  accomplished 
by  drawing  the  normal  curve  in  its  correct  relation  to 
the  line  F  G,  as  shown  by  N"  O"  P",  when  the  raking 


88 


M<t«l 


process    could   be   reversed,    thereby  developing    the 
curve  O  N  P  one-half  of  the  plan  of  top. 

Preparatory  to  obtaining  the  varying  width  of  the 
pattern  of  the  side,  a  number  of  points  must  be  fixed 
upon  in  the  curves  of  both  top  and  bottom  from  which 
to  take  the  measurements.  As  one-quarter  of  the  top 
is  already  divided  into  spaces,  another  quarter,  0  P, 
may  be  divided  into  the  same  number  of  spaces  (also 
dividing  0"  P"  into  the  same  space  as  0"  N").  If 
N"  O''  P"  is  the  normal  curve  of  the  top  it  would  very 
naturally  be  divided  into  equal  spaces  by  the  dividers, 
as  is  usual  in  such  cases,  while  the  spacing  in  N  O  P 
would  be  the  result  of  the  operation  of  raking.  It  is 
advisable  to  have  the  spaces  in  N"0"P"  all  equal  to 
each  other,  as  it  is  from  this  curve  that  the  stretch-out 
of  the  top  of  the  pattern  is  to  be  derived,  the  conven- 
ience of  which  will  become  apparent  when  the  pattern 
is  developed. 

The  quarter  O  P  is  used  in  connection  with  the  quar- 
ter O  N,  because  these  two  combined  constitute  a  half  of 
the  top  curve  lying  on  one  side  of  the  line  A  C  of 
the  plan  which  divides  the  article  into  symmetrical 
halves,  it  being  only  necessary,  when  the  shape  of  an 
article  permits,  to  obtain  the  pattern  of  one-half  and 
then  to  duplicate  by  any  convenient  means  to  obtain  the 
other  half. 

The  corresponding  half  of  the  plan  of  the  base, 
therefore,  ABC,  must  also  be  divided  into  the  same 
number  of  equal  spaces  as  were  used  at  the  top,  all  as 
shown  in  the  drawing,  and  both  sets  of  points  should 
be  numbered  alike,  beginning  at  the  same  side. 

Having  thus  fixed  the  points  from  which  measure- 
ments across  the  pattern  of  the  side  are  to  be  taken, 
next  draw  lines  across  the  plan  connecting  points  of 
like  number,  as  shown  by  the  full  lines  in  the  plan. 
This  divides  the  entire  side  of  the  article  into  a  num- 
ber of  four-sided  figures;  but  as  it  is  necessary,  as 
shown  above,  to  have  it  divided  into  triangles,  each 
four-sided  figure  may  now  be  redivided  by  a  line  drawn 
through  its  opposite  angles,  thus  cutting  it  into  two 
triangles.  In  other  words,  each  point  in  the  base 
should  bo  connected  with  a  point  of  the  next  lower 
number  (or  higher,  according  to  circumstances)  in  the 
curve  of  the  top,  and  these  lines  should  be  dotted  in- 
stead of  full  lines  for  the  sake  of  distinction  and  to 
avoid  confusion  in  subsequent  parts  of  the  work. 
Thus  1  of  the  base  is  connected  by  a  dotted  line  with 
0'  of  the  top,  2  of  the  base  with  1'  of  the  top,  etc. 

In  respect  to  which  is  the  best  way  to  run  the 
dotted  lines,  common  sense  will  be  the  best  guide. 


Thus,  in  the  space  bounded  by  the  lines  4  4'  and  5  5', 
it  is  plainly  to  be  seen  that  there  would  be  greater  ad- 
vantage and  less  liability  of  error  in  connecting  .">  of  the 
bottom   curve  with   4'  of  the  top  than  in  crossing  the 
I  line  from  -i  of  the  bottom  to  5'  of  the  top,  for  the  rea- 
;   son  that  in   the  former    case    the    triangles   produced 
would  be  less  scalene  or  acute. 

The  next  step  is  to  devise  a  means  of  determining 
the  true  lengths  which  these  lines  represent  or,  in  other 
words,  their  real  length  as  they  could  be  measured  if  a. 
full  size  model  of  the  article  were  cut  from  a  block  of 
wood  or  clay  upon  which  these  lines  had  been  marked, 
as  shown  upon  the  drawing. 

The  lines  upon  the  plan,  of  course,  "only  show  the 
horizontal  distances  between  the  points  which  they 
connect.  The  vertical  hight  above  the  base  of  any  of 
the  points  in  the  upper  curve  can  easily  be  found  by 
measuring  from  its  position  upon  the  line  F  G  of  the 
elevation  perpendicularly  to  the  base  E  H.  Therefore, 
having  both  the  vertical  and  the  horizontal  distance 
given  between  any  two  points,  it  is  only  necessary  to 
construct  with  these  dimensions  a  right  angle  triangle, 
and  the  hypothenuse  will  give  their  true  distance  apart. 
Thus  in  Fig.  262  a  b  is  equal  to  the  line  4  4'  of  the 
plan,  while  a  c  is  made  equal  to  4  4'  of  the  elevation. 
Consequently  c  b  represents  the  true  distance  between 
the  points  4'  of  the  top  and  4  of  the  base.  Therefore, 
to  obtain  all  of  these  hypothenuses  in  the  simplest 
possible  manner,  it  will  be  necessary  to  construct  one 
or  two  diagrams  of  triangles.  To  avoid  confusion  it 
is  better  to  make  two ;  one  for  obtaining  the  distances 
represented  by  the  full  or  solid  lines  drawn  across  the 
plan  and  the  other  for  those  of  the  dotted  lines.  To 
do  this  extend  the  base  line  E  II  of  the  elevation,  as 
shown  at  the  left,  at  any  convenient  points,  in  which, 
as  R  and  S,  erect  two  perpendicular  lines.  Project 
lines  horizontally  from  all  the  points  in  F  G,  cutting 
these  two  lines  as  shown,  and  number  the  points  of 
intersection.  (Some  of  the  figures  are  omitted  in  the 
drawing  for  lack  of  space.)  From  R  set  off  on  the  base 
line  distances  equal  to  the  lengths  of  the  solid 
lines  of  the  plan  1  1',  2  2',  3  3',  etc.,  numbering  the 
points!,  2,  3,  etc.,  as  shown,  and  connect  points  of 
similar  number  upon  the  base  with  those  upon  the 
perpendicular.  From  S  set  off  on  the  base  line  dis- 
tances equal  to  the  lengths  of  the  dotted  lines  of  the 
plan  1  0',  2  1',  3  2',  etc.,  and  number  them  to  corre- 
spond with  figure  upon  the  line  of  the  base  ABC. 
Thus  make  1  S  equal  to  1  0'  of  the  plan,  2  S  equal  to 
2  1'  of  the  plan,  3  S  equal  to  3  2',  etc.,  and  connect 


Pri. 


of  Puiii  m    < 'nit!  in/. 


eacli  point  in  tlie  base  with  the  point  of  next  lower 
number  upon  the  perpendicular  by  ;i  dotted  line,  ;is  1 
<>n  tin-  base  with  ii  nu  the  perpendicular,  2  with  1,  3 
with  2,  etc.  The  entire  surface  of  the  piece  for  which 


Fig.  £63.— Top,  Back  and  Bottom  for  a  Model  of  One-half  the 
Article  Shown  in  Fig.  201. 

a  pattern  is  required  has  thus  been  cut  up  into  two  sets 
of  triangles,  one  set  having  the  spaces  upon  the  base 
line  ABC,  which  arc  all  equal,  for  their  bases,  and 
the  other  set  having  the  spaces  in  the  curve  N"  0"  P" 
of  the  top,  also  equal  to  each  other,  as  their  bases,  and 
each  separate  triangle  having  one  solid  line  and  one 
dotted  line  as  its  sides. 

In  all  of  this  work  the  student's  powers  of  mental 
conception  are  called  into  play.  The  shape  of  the  sur- 
face, which  is  yet  to  be  developed,  has  been  spoken  of 
as  if  it  really  existed — in  fact,  it  must  exist  in  the  mind 
or  imagination  of  the  operator  in  order  to  make  him 
intelligent  as  to  what  he  is  doing.  If  this  fails  him, 
he  can  resort  to  a  model  which  can  easily  be  con- 
structed (full  size  or  to  scale,  according  to  convenience) 
as  follows :  Describe  upon  a  piece  of  cardboard  or 
metal  the  shape  E  F  G  II,  Fig.  261,  to  which  add  on 
its  lower  side,  E  H,  one-half  of  the  plan  of  the  bottom, 
ABC,  with  the  curve  N  0  P  and  the  solid  lines  con- 
necting it  with  the  outside  curve  traced  thereon.  Also 
add  on  its  upper  side,  F  G,  one-half  the  shape  of  top, 
N"  0"  P",  marking  the  points  1,  2,  3,  etc.,  upon  its 
edge.  Now  cut  out  the  entire  shape  in  one  piece,  as 
shown  in  Fig.  263,  and  bend  the  same  at  right  angles, 
on  the  lines  F  G  and  E  H.  Small  triangles  of  the 
shape  and  size  of  each  of  the  triangles  shown  in  the 
diagram  of  solid  lines,  Fig.  261,  as  0  0  K,  1  1  K,  2  2  E, 


etc.,  can  now  be  cut  out  and  placed  upon  the  portion 
representing  the  bottom,  each  with  its  base  upon  the 
solid  line  which  it  represents,  at  the  same  bringing  the 
apex  of  each  to  the  corresponding  number  on  the  top. 
These  can  be  fastened  in  place  by  bits  of  sealing 
wax,  or  if  cut  from  metal  the  whole  can  be  soldered 
together. 

The  hypothenuses  of  the  various  triangles  will 
thus  represent  the  true  distances  across  the  pattern 
upon  the  solid  lines  of  the  plan,  while  the  distances 
upon  the  dotted  lines  can  be  represented  by  pieces  of 
thread  or  wire,  placed  so  that  each  will  reach  from  the 
point  at  the  base  of  one  of  the  triangles  to  the  point  at 
the  top  of  the  one  next  it.  If  constructed  of  metal 
two  or  three  triangles  will  suffice  to  give  the  model 
sufficient  rigidity,  and  the  remaining  points  can  be 
connected  by  pieces  of  wire,  using  a  different  kind  of 
wire  to  represent  the  distances  on  the  dotted  lines. 

In  Fig.  264,  is  shown  a  pictorial  representation  of 
a  model  constructed,  as  above  described,  from  the  draw- 
ings shown  in  Fig.  261.  In  the  illustration  the  tri- 
angles 2,  5  and  8  only  are  shown  in  position,  their 
hypothenuses  connecting  points  of  similar  number  in 
the  upper  and  lower  bases.  The  other  points  are  rep- 
resented as  being  connected  by  wires  or  threads  repre- 
senting both  the  solid  and  the  dotted  hypothenuses  in 
the  diagrams  of  triangles  in  Fig.  261.  Such  a  model  if 
constructed  will  give  a  general  idea  of  the  shape  of  the 
entire  covering,  and  at  the  same  time  £>f  the  small 
pieces,  or  triangles,  of  which  the  covering  is  composed, 
with  all  the  dimensions  of  each.  If  all  of  the  spaces 
formed  upon  this  skeleton  surface  could  be  filled  in 


Fig.  264.— Perspective  View  of  Cardboard  Model  of  One-half  the 
Article  Shown  in  Fig.  261. 


Avith  pieces  of  cardboard  or  metal  just  the  size  of  each 
and  the  whole  removed  together  and  flattened  out  (each 
piece  being  fastened  to  its  neighbor  at  the  sides),  it 
would  constitute  the  required  pattern,  the  same  as  will 
be  subsequently  obtained  by  measurements  taken  from 
the  drawing,  and  as  shown  in  Fig.  265. 


90 


Tlir   Xrtr    Mi-tal     HW.vr    1'ntln-n    Ifovl: 


Having  by  means  of  the  diagrams  of  triangles  in 
Fig.  261  obtained  the  lengths  of  all  the  sides  it  is  now 
only  necessary  to  construct  successively  eaeh  triangle 
in  the  manner  described  in  Chapter  IV.  Problem  36, 
remembering  that  the  last  long  side  of  each  triangle 
used  is  also  the  first  long  side  of  the  next  one  to  be 
constructed.  Therefore,  at  any  convenient  place  draw 
any  straight  line,  A  N  of  Fig.  265,  which  make  equal 
to  the  real  distance  from  A  to  N,  Fig.  261,  which  has 
been  found  to  be  the  distance  0  0  of  the  diagram  of 
solid  lines.  To  conduct  this  operation  with  the  great- 
est economy  and  ease  it  is  necessary  to  have  two  pairs 
of  dividers,  which  shall  remain  set,  one  to  the  spaces 
upon  the  plan  of  the  base  ABC,  and  the  other  to  tin- 
spaces  upon  ~S"  O"  P",  and  a  third  pair  for  use  in 
taking  varying  measurements.  From  A  of  Fig.  265 
as  a  center,  with  a  radius  equal  to  0  1  of  the  plan,  Fig. 
261,  describe  a  small  arc,  and  from  N  as  a  center,  with 
a  radius  equal  to  the  true  distance  from  N  to  1  of  the 
plan,  which  has  been  found  to  be  0  1  of  the  diagram 
of  dotted  lines,  describe  another  arc,  cutting  the  first 
one  as  shown  at  the  point  1,  Fig.  265.  The  triangle 
thus  constructed  represents  the  true  dimensions  of  one 
indicated  by  the  same  figures  of  the  plan.  Next  from 
N  of  the  pattern  as  a  center,  with  a  radius  equal  to 
N"  1  of  true  profile  of  top,  Fig.  261,  describe  a  small 
arc,  which  cut  with  one  struck  from  point  1  of  pat- 
tern as  a  center,  with  a  radius  equal  to  1  1  of  the  dia- 
gram of  solid  lines,  thus  locating  point  1'  of  pattern. 
This  triangle  is,  in  turn,  succeeded  by  another  whose 
sides  are  next  in  numerical  order,  that  is  1  2  of  the  base 
and  1  2  of  the  diagram  of  dotted  lines.  Thus  the 
operation  is  continued,  always  letting  the  spaces  of  the 
circumference  of  base  succeed  one  another  at  one  side 
of  the  pattern,  and  the  spaces  upon  the  true'profile  of 
top  succeed  one  another  at  the  other  side  of  the  pat- 
tern, until  all  the  triangles  have  been  laid  out  as 
shown  by  A  N  P  C,  Fig.  265,  which  will  complete 
one-half  the  entire  pattern. 

It  is  not  necessary  to  draw  all  of  the  dotted  or 
solid  lines  across  the  pattern,  as  the  points  where  the 
small  arcs  intersect  are  all  that  are  really  needed  in 
obtaining  the  outlines  of  the  pattern,  but  it  is  often 
advisable  to  draw  them  as  well  as  to  number  each  new 
point  as  obtained,  in  order  to  avoid  confusion  and 
insure  the  order  of  succession. 

In  dividing  the  curves  of  top  and  bottom  into 
spaces,  such  a  number  of  points  should  be  taken  as 
will  insure  the  greatest  accuracy,  as  in  the  case  of 
dividing  a  profile.  Thus  too  few  would  give  too  short 


a  stretch-out,  while  if  the  spaces  were  too  small  error  in 
transferring  their  lengths  might  result,  which  would  be 
increased  as  many  times  as  there  were  spaces. 

I  nder  the  head  of  transition  pieces  may  be  in- 
cluded a  large  number  of  forms  having  various  shaped 
polygonal  or  curved  figures  as  their  upper  and  lower 
surfaces,  placed  at  various  angles  to  each  other,  some- 
times centrally  located  as  they  appear  upon  the  plan 
and  sometimes  otherwise.  It  often  happens  that  one 
surface  or  termination  is  entirely  outside  the  other  in 
that  view,  forming  an  offset  between  pipes  of  differing 
sixes  and  shapes.  Sometimes  such  an  offset  takes  a 
curved  form,  constituting  a  curved  elbow  of  varying 
section  throughout  its  length,  in  which  case  it  consists 
of  a  number  of  pieces,  each  with  a  different  shape  at 
either  end.  With  such  forms  may  be  classed  the  ship 


Fig.  265.— One-half  Pattern  of  Side  of  Article  Shmvn  in  Fig.  261. 

ventilator,  whose  lower  end  is  usually  round  and  hori- 
zontal and  whose  upper  end  is  enlarged  and  elliptical 
and  stands  in  a  vertical  position,  the  whole  being  com-: 
posed  of  five  or  six  pieces.  In  such  cases,  when  the 
shape  and  position  of  the  two  terminating  surfaces  onlv 
are  given,  it  becomes  necessary  to  assume  or  draw  as 
many  intermediate  surfaces  as  there  are  joints  required, 
each  of  such  a  shape  that  the  whole  series  will  form  a 
suitable  transition  between  two  extreme  shapes.  It, 
may  be  remarked,  that  what  have  been  spoken  of 
here  as  "surfaces"  do  not  necessarily  mean  surfaces 
of  metal  forming  solid  ends  to  the  pieces  describe) I. 
but  simply  outlines  upon  paper  to  work  to,  as  more 
often  the  "surface"  is  really  an  opening. 

Still  another  class  of  forms  demanding  treatment 
by  triangulation  result,  from  the  construction  of  arches 


Princijlles   of  Pntli-rn    Cnttinrj. 


91 


cut  through  ciirvwd  walls,  as  when  an  arch  of  either 
round  or  elliptical  form,  as  a  door  or  window  head,  is 
placed  in  a  circular  wall  in  such  a  manlier  1  hut  its  sides 
or  jambs  are  radial,  or  tend  toward  the  center  of  the 
curve  of  the  wall.  It  will  l>e  seen  that  the  sofht  of 
such  an  arch  is  similar  in  shape  to  the  sides  of  a  transi- 
tion piece,  having  what  might  be  called  its  upper  and 
lower  surfaces  curved  and  placed  vertically.  In  such 
cases  it.  is  best  to  consider  the  horizontal  plain;  passing 
through  the  springing,  line  of  the  arch  as  the  base  from 
which  to  measure  the  hights  of  all  points  assumed  in 
the  outer  and  inner  curves. 

It  is  believed  that  a  sufficient  number  of  this  gen- 


Fig.  S66. — Elevations  and  Plan  of  an  Elliptical  Cone. 

eral  class  of  problems  will  be  found  in  the  third  section 
of  the  chapter  on  Pattern  Problems  to  enable  the  care- 
ful student  to  apply  the  principles  here  explained  to 
any  new  forms  that  might  present  themselves  for  his 
consideration,  remembering  that  any  form  may  be  so 
turned  as  to  bring  any  desired  side  into  a  horizontal 
position  to  be  used  as  a  base,  or  that  an  upper  hori- 
zontal surface  can  be  used  as  a  base  as  well  as  a  lower. 
The  operations  of  triangulation  undoubtedly  re- 
quire more  care  for  the  sake  of  accuracy  than  those  of 
any  other  method  of  pattern  cutting,  for  the  reason  that 
there  is  no  opportunity  of  stepping  off  a  continuous 
stretchout,  at  once,  upon  any  line,  either  straight  or 
curved.  It  is  therefore  not  to  be  recommended  if  the 


subject  in  hand  admits  of  treatment  by  any  regular 
method  without  too  much  subdivision.  Triangulation 
is  not  introduced  as  an  alternate  method,  but  as  a  last 
resort,  when  nothing  else  will  do. 

Besides  the  various  forms  of  transition  pieces,  an- 
other class  of  forms  is  to  be  treated  under  this  head, 
which  might  almost  be  considered  as  regular  tapering 
articles.  They  include  shapes,  or  frustums  cut  from 
shapes,  which  terminate  in  an  apex,  but  whose  bases 
cannot  be  inscribed  in  a  circle,  as  irregular  polygons, 
figures  composed  of  irregular  curves  as  well  as  the  per- 
fect ellipse.  A  solid  whose  base  is  a  perfect  ellipse 
and  whose  apex  is  located  directly  over  the  center  of 
its  base  (in  other  words,  an  elliptical  cone)  is  perhaps 
the  best  typical  representative  of  this  class  of  figures. 
If  the  base  of  such  a  cone  be  divided  into  quarters  by 
its  major  and  minor  axes,  it  will  be  seen  at  once  that 
all  of  the  points  in  the  perimeter  of  any  one  quarter 
will  be  at  different  distances  from  the  apex  of  the  cone, 
because  they  are  at  different  distances  from  the  center 
of  base  or  the  intersection  of  the  two  axes.  This  is 
clearly  shown  in  Fig.  266,  in  which  are  shown  the  two 
elevations  and  the  plan  of  an  elliptical  cone.  The  side 
elevation  shows  K  E  to  be  the  distance  of  the  apex 
from  the  point  P  in  the  plan  of  the  base,  while  the  end 
elevation  shows  K'  D  to  be  the  distance  of  the  apex 
from  the  point  D  of  the  base,  or  the  true  distance  rep- 
resented by  X  D  of  the  plan. 

If  one-quarter  of  the  plan  of  the  base,  as  D  P,  be 
divided  into  any  convenient  number  of  equal  spaces 
and  lines  be  drawn  to  the  center  X,  as  shown,  each  line 
will  represent  the  horizontal  distance  of  a  point  in  the 
perimeter  from  the  apex ;  and  if  a  section  of  the  cone 
be  constructed  upon  any  one  of  these  lines,  as,  for  in- 
stance, line  4  X,  or,  in  other  words,  if  a  right  angle 
triangle  be  drawn,  of  which  4  X  is  the  base  and  E  K 
the  altitude,  the  hypothenuse  will  be  the  true  distance 
of  the  point  4  from  the  apex.  Therefore,  to  ascertain 
the  distances  from  the  apex  to  the  various  points  in  the 
circumference  of  the  base  construct  a  simple  diagram 
of  triangles,  as  shown  in  Fig.  2G7,  viz.  :  Erect  any 
perpendicular  line,  as  X  M,  equal  in  hight  to  E  K 
of  the  elevation;  from  X,  on  a  horizontal  line  X  P,  as 
a  base,  set  off  the  various  distances  of  the  plan,  X  1, 
X  2,  X  3,  etc.,  numbering  each  point,  and  from  each 
point  draw  a  line  to  M.  These  hypothenuses  will  then 
represent  the  distances  of  the  various  points  in  the 
perimeter  of  the  base  from  the  apex  of  the  cone;  or,  in 
other  words,  the  sides  of  a  number  of  triangles  forming 
the  envelope  of  the  cone,  the  bases  of  which  triangles 


92 


Tlic  \i'n- 


f'ufti-rn  Book. 


will  be  the  spaces  1  2,  2  3,  etc.,  upon  the  plan.  As  all 
of  these  triangles  terminate  at  a  common  apex  or 
center,  instead  of  laying  out  each  one  separately  to 
form  a  pattern,  as  in  the  case  of  an  article  of  the  type 
shown  in  Fig.  261,  the  simplest  method  is  as  follows : 


Fig.  S67. — Diagram  of  Sections  on  the  Radio]  Lines  of  the  Plan  in 
/•'('</.  260,  to  which  in  Added  the  Pattern  of  One-quarter  of  the 
Envelope. 

From  M,  of  Fig.  267,  as  a  center,  with  radii  corre- 
sponding to  the  distances  from  M  to  points  on  P  X,  as 
M  1,  M  2,  M  3,  etc.,  describe  arcs  indefinitely,  as 
shown  to  the  left;  then  taking  the  space  used  in  step- 
ping off  the  plan  between  the  points  of  the  dividers, 
place  one  foot  upon  the  arc  drawn  from  point  8,  as  at 
D,  and  swing  the  other  foot  around  till  it  cuts  the  arc 
drawn  from  point  7;  from  this  intersection  as  a  center 
swing  it  around  again,  cutting  the  arc  from  6;  or  in 
other  words,  step  from  one  arc  to  the  next  till  one- 
quarter  of  the  circumference  has  been  completed. 

As  the  spaces  in  the  base  are  equal,  it  is  clearly  a 
matter  of   convenience  whether  this  last  operation    is 


Fig.  268.— Frustum  of  an  Elliptical  Cone. 

begun  upon  arc  8,  stepping  first  to  arc  7,  then  to  arc 
6,  etc.,  or  whether  it  is  begun  upon  arc  1,  stepping 
first  to  arc  2,  then  to  3,  etc.,  till  complete.  A  line 
traced  through  these  points,  as  A  D,  will  give  the  cut 


at  the  base  of  the  envelope,  and   A   P  M  will   be  the 
envelope  of  one-quarter  of  the  cone. 

In  Fig.  26S  is  shown  a  perspective  view  of  the 
frustum  of  the  cone  shown*  in  Fig.  26f>,  the  upper  sur- 
face A  B  being  shown  in  Fig.  266  by  the  lines  G  II 
and  N  O.  If  the  envelope  of  such  a  frustum  is  desired 
the  cut  which  its  upper  surface  would  make  through 
the  envelope  of  the  entire  cone  could  lie  obtained  in 
exactly  the  same  manner  as  that  of  its  lower  base,  be- 
cause the  upper  surface  of  the  frustum  is  in  reality  the 
base  of  the  cone,  which  remains  above  after  the  lower 
part  has  been  cut  away.  But  as  part  of  the  operation 
has  already  been  performed  in  obtaining  the  cut  at  the 
base,  it  is  most  easily  accomplished  as  follows :  First 
draw  radial  lines  from  the  point  M  of  the  diagram  of 


Fig.  S6S. — Elevation  of  the  Frustum  of  an  Elliptical  Cone. 

triangles,  Fig.  267,  to  each  of  the  points  previously 
obtained  in  the  cut  at  the  bottom  of  the  envelope, 
between  A  and  D;  also  draw  a  horizontal  line  at  a 
hight  above  the  base  X  P  equal  to  R  L,  Fig.  266,  cut- 
ting the  hypothennses  M  1,  M  2,  etc.,  as  shown  by 
G  H.  Now  place  one  foot  of  the  dividers  at  the  point 
M,  and  bringing  the  other  foot  successively  to  the 
various  points  of  intersection  of  the  liiie  G  II  with  the 
various  hypothenuses,  describe  arcs  cutting  the  radial 
lines  in  the  envelope  of  corresponding  number.  A  line 
traced  through  the  points  of  intersection,  as  B  C,  will 
give  the  cut  at  the  top  of  the  envelope  of  the  frustum, 
of  which  A  D  is  the  bottom  cut. 

If  the  cut  at  the  top  of  the  frustum  is  to  be  oblique 
instead  of  horizontal,   a  means  must  be  devised   for 


of  Pattern  Cutting. 


measuring  the  distance  from  the  apex  at  which  the 
oblique  plane  cuts  each  of  the  hypothennses,  or  in  otlier 
words,  eacli  of  the  lines  drawn  from  the  apex  of  the 
cone  to  the  various  points  in  its  liase.  In  Fig.  269, 
K  S  T  F  is  the  elevation  of  an  oblique  frustum  of  an 
elliptical  cone,  whose  apex  is  at  K,  and  whose  base  is 
the  same  and  lias  been  divided  in  the  same  manner  as 
that  shown  in  Fiji.  266. 

Erect  lines  from  each  of  the  points  iu  the  curve  of 
one-half  the  plan  P  1)  A  to  the  base  line  K  F  of  the 
elevation,  thence  carrv  them  toward  the  apex  K,  cut- 
ting the  line  S  T  ;  the  vertical  liight  of  the  points  upon 
S  T  can  then  most  easilv  lie  measured  1>\-  carrying  them 
horizontally,  cutting  the  center  line  R  K  of  the  cone, 
where  to  avoid  confusion  they  should  be  numbered  to 


Fig.  270. — Diagram  of  Sections  on  the  Radial  Lines  of  the  Plan  in 
Fig.  269,  with  the  Pattern  of  One-half  the  Envelope. 

correspond  with  the  points  of  the  plan  from  which  each 
was  derived.  These  points  may  now  be  transferred  in 
a  body  by  any  convenient  means  to  the  vertical  line 
X'  M'  of  the  diagram  of  triangles,  Fig.  270,  seeing  that 
each  point  is  placed  at  the  same  distance  from  M'  that 
it  is  from  the  point  K  of  Fig.  269.  A  horizontal  line 
from  any  one  of  the  points  on  the  line  X'  M'  extended 
to  the  hvpothenuse  of  corresponding  number  will  then 
give  the  correct  distance  of  that  point  from  the  apex 
cf  the  cone.  The  diagram  M'  X'  I)'  is  a  duplicate  of 
M  X  P  of  Fig.  267,  and  the  lower  outline  of  the  en- 
velope is  the  same  as  that  shown  in  Fig.  267.  It  will 
be  noted,  however,  that  half  the  stretchout  of  the  base 
is  iiecessarv  in  this  case  to  give  all  the  essentials  of  the 
pattern  of  the  envelope,  while  one-quarter  was  suiiicient 
for  the  previous  operations.  When  all  the  points  in 
the  uppei  line  of  the  frustum  have  been  obtained  in  the 


diagram  they  may  be  transferred  to  the  various  radial 
lines  in  the  envelope,  from  M'  as  a  center,  by  the  use 
of  the  compasses  as  before,  all  as  shown  in  the  draw- 
ing. 

If  the  apex  of  the  cone  were  not  located  directly 
over  the  crossing  of  the  two  axes  of  the  ellipse — that  is, 
if  the  cone  were  scalene  or  oblique  instead  of  right — 
the  method  of  obtaining  its  "envelope,  or  parts  of  the 
same,  would  not  differ  from  the  foregoing.  Lines 
drawn  from  the  points  of  division  in  the  circumference 
of  the  base  to  the  point  representing  the  position  of 
the  apex  in  the  plan  will  lie  the  horizontal  distances 
used  in  constructing  a  diagram  of  triangles,  which  dis- 
tances can  be  used  in  connection  with  the  vertical 
hight  of  the  cone,  as  before,  in  obtaining  the  various 
hypothenuses.  If  the  apex  of  a  scalene  cone  be 
located  over  the  line  of  either  axis  of  the  ellipse,  either 
within  the  perimeter  of  the  base  or  upon  one  of  those 
lines  continued  outside  the  base,  one-half  the  pattern 
of  the  entire  envelope  will  have  to  be  obtained  at  one 
operation ;  but  if  the  apex  is  not  located  upon  either 
of  those  lines  in  the  plan,  then  the  entire  envelope 
must  be  obtained  at  one  operation,  as  no  two  quarters 
or  halves  of  the  cone  will  be  exactly  alike. 

The  method  of  obtaining  the  envelope  of  any 
scalene  cone,  even  though  its  base  be  a  perfect  circle, 
is  governed  by  the  same  principles  as  those  employed 
in  the  above  dernonsti-ations. 

It  will  be  well  to  remember  that  any  horizontal 
section  of  a  scalene  cone  is  the  same  shape  as  its  base, 
which  fact  can  be  used  to  ad  vantage  in  determining  the 
best  method  to  be  employed  in  obtaining  the  envelope 
of  any  irregular  flaring  surface  that  may  be  presented. 
If,  for  instance,  the  plan  of  any  article,  whose  upper 
and  lower  surfaces  are  horizontal,  shows  each  to  con- 
sist of  two  circles  or  parts  of  circles  of  different  diame- 
ters not  concentric,  it  is  evident  that  the  portion  of  the 
envelope  indicated  by  the  circles  of  the  plan  is  part  of 
the  envelope  of  a  scalene  cone.  An  illustration  of  this 
is  given  in  Fig.  271,  which  shows  a  portion  of  an 
article  having  rounded  corners  and  flaring  sides  and 
ends,  but  with  more  flare  at  the  end  than  at  the  side. 
The  plan  shows  the  curve  of  the  bottom  corner  A  B  to 
be  a  quarter  circle  with  its  center  at  X,  and  that  of  the 
top  C  D  to  be  a  quarter  circle  with  its  center  at  Y. 
The  rounded  cornet  A  B  I)  C  is  then  a  portion  of 
the  envelope  of  a  frustrum  of  a  scalene  cone,  and  the 
method  of  finding  the  dimensions  of  the  complete  cone 
is  quite  simple  and  is  as  follows:  First  draw  a  line, 
Z  N,  through  the  centers  of  the  two  circles  in  the  plan, 


T/ie  Xew  Metal    Worker  Pattern  Book. 


at  right  angles  to  which  project  an  oblique  elevation,  as 
shown  below,  making  the  distance  between  the  two 
lines  E  F  and  G  H  equal  to  the  hight  of  the  article. 
Lines  from  X  arid  M  of  the  plan  of  the  bottom  fall 
upon  G  H,  locating  the  points  X'  and  H,  while  lines  ! 
from  Y  and  N  of  the  top  locate  the  points  Y'  and  F  in 
the  upper  line  of  the  oblique  elevation.  A  line  drawn 
through  Y'  and  X',  the  centers  of  the  circles,  will  then 
represent  the  axis  of  the  cone  in  elevation,  which  can 
be  continued  to  meet  a  line  drawn  through  the  points 
F  and  H,  representing  the  side  of  the  cone,  thus  locat- 
ing the  apex  Z'  of  the  scalene  cone.  The  point  Z'  can 
then  be  carried  back  to  the  plan,  as  shown  at  Z,  thus 
locating  the  apex  in  that  view.  As  the  line  N  Z  rep- 
resents the  horizontal  distance  between  the  point  F  and 
the  apex  Z'  of  the  cone,  so  lines  drawn  from  Z  to  any 
number  of  points  assumed  in  the  curve  of  the  base 
C  D  will  give  the  horizontal  distances  between  those 
points  and  the  apex,  to  be  used  as  the  bases  in  a  dia- 
gram of  triangles  similar  to  that  shown  in  Fig.  267, 
while  V  Z'  gives  their  hight.  Having  drawn  a  dia- 
gram of  triangles  the  pattern  follows  in  the  manner 
there  shown. 

For  greater  accuracy  in  the  case  of  a  very  tapering 
cone,  the  circles  of  the.  plan  can  be  completed,  as 
shown  dotted,  and  their  points  of  intersections  with  the 
line  Z  N  can  be  dropped  into  oblique  elevation,  as 
seen  at  S  arid  T,  through  which  a  line  can  be  drawn  to 
meet  a  line  through  F  and  H  with  greater  accuracy 
than  one  through  Y'  and  X',  as  the  angle  in  the  former 
case  is  twice  as  great. 

In  the  above  methods  of  obtaining  the  envelopes 
of  what  may  be  termed  irregular  conical  forms,  it  will 
be  clearly  seen  that  the  operation  of  dividing  the  curve 
of  the  base  into  a  great  number  of  spaces  really  re 
solves  the  conical  figure  into  a  many  sided  pyramid, 
and  that  the  lines  connecting  the  apex  with  the  points 
in  the  base,  which  have  been  referred  to  as  hypothe- 
n  uses,  are  really  the  angles  or  hips  of  the  pyramid. 
It  is  therefore  self  evident  that  any  method  of  de- 
velopment which  is  applicable  to  a  many  sided  pyra- 
mid is  equally  applicable  to  one  whose  sides  are  fewer 
in  number,  with  the  only  difference,  however,  that  the 
lines  representing  the  angles  or  hips  in  the  case  of  a 
pyramidal  figure  mean  angles  or  sharp  bends  in  the 
pattern  of  the  envelope,  while  in  the  case  of  the  conical 
envelope  the  bends  arc  so  slight  as  to  mean  only  a  con- 
tinuous form  or  curve. 

It  is  believed  that  the  foregoing  elucidation  of  the 
principles  governing  the  development  of  the  surfaces 


of  irregular  shaped  figures  is  sufficiently  clear  to  make 
the  demonstrations  of  this  class  of  problems,  given  in 
Chap.  VI,  Section  3,  easily  understood  by  the  student, 
as  well  as  to  enable  him  to  apply  them  to  any  new 
forms  that  may  present  themselves  for  solution. 

This  chapter  is  intended  to  present,  under  its  three 
different  heads,  all  the  principles  neccssarv  to  guide 
the  student  in  the  solution  of  any  problem  that  may 


Fi(j.  ~'71. — Elevations  and  Plan  of  an  Article   the    Corner  of  which 
is  a  Portion  of  a  Scalene  Cone. 


arise.  Its  aim  is  to  teach  principles  rather  than  rules, 
and  the  student  is  to  be  cautioned  against  arbitrary 
rules  and  methods  for  which  he  cannot  clearly  under- 
stand the  reason.  His  good  sense  must  govern  him 
in  the  employment  of  principles  and  in  the  choice 
of  methods.  There  is  hardly  a  pattern  to  be  cut  which 
cannot  be  obtained  in  more  than  one  way.  Under 
some  conditions  one  method  is  best,  and  under  other 
conditions  another,  and  careful  thought  before  the 


of  Pattern    Outliu*/. 


95 


drawing  is  begun  will  show  which  is  best  for  the  pur- 
pose in  hand. 

The  list  of  problems  and  demonstrations  "in   the 
rhaptt-r  which  follows  is  believed  to  be  so  comprehen- 


sive that  therein  will  be  found  a  parallel  to  almost  any- 
thing that  may  be  required  of  the  pattern  cutter,  and 
it  is  believed  that  he  will  have  no  difficulty  in  applying 
them  to  his  wants. 


CHAPTER  VI. 


Every  effort  has  been  put  forth  in  the  preceding 
chapters  of  this  book  to  prepare  the  student  for  the  all 
important  work  which  is  to  follow — viz.,  the  solution 
of  pattern  problems.  It  is  always  advisable  in  the 
study  of  any  subject  to  be  well  grounded  in  its  funda- 
mental principles.  For  this  reason  a  chapter  on  Linear 
Drawing  has  been  prepared  to  meet  the  requirements  of 
the  student  in  pattern  cutting,  which  is  preceded  by  a 
description  of  drawing  materials  and  followed  bv  a 
solution  of  the  geometrical  problems  of  most  frequent 
occurrence  in  his  work.  But  the  most  important 
chapter  is  the  one  immediately  preceding  this,  in  which 
the  theory  of  pattern  cutting  is  explained,  and  which, 
if  thoroughly  understood,  will  render  easy  the  solution 
of  any  problem  tlic  student  may  chance  to  meet. 

The  selection  of  problems  here  presented  is  made 
sufficiently  large  and  varied  in  character  to  anticipate, 
so  far  as  possible,  the  entire  wants  of  the  pattern  cut- 
ter, and  the  problems  are  so  arranged  as  to  be  con- 
venient for  reference  by  those  who  make  use  of  this 


part  of  the  book  without  previous  study  of  the  other 
chapters. 

Tn  the  demonstrations,  onlv  the  scientific  phase  of 
the  subject  will  be  considered;  consequently,  all  al- 
lowances for  seams,  joints,  etc.,  as  well  as  determining 
where  joints  shall  be  made,  arc  at  the  discretion  of 
the  workman.  In  some  of  the  problems  it  has  been 
necessary  to  assume  a  place  for  a  joint,  but  if  the  joint 
is  required  at  a  place  other  than  where  shown,  the 
method  of  procedure  would  be  slightly  varied  while 
the  principle  involved  would  remain  the  same. 

Each  demonstration  will  be  complete  in  itself.  Al- 
though references  to  other  problems,  principles,  etc., 
will  be  made  where  such  references  will  be  of.advnn- 
tage  to  the  student. 

As  stated  in  the  preceding  chapter,  the  problems 
will  be  classed  under  three  different  heads  according 
to  the  forms  which  thev  embodv — viz.  :  First, 
Parallel  Forms;  Second.  Regular  Tapering  Forms,  and 
Third,  Irregular  Forms. 


SECTION   1. 


(MITER   CUTTING). 


The  problems  given  in  this  section  are  such  as 
occur  in  joining  moldings,  pipes  and  all  regular  con- 
tinuous forms  at.  any  angle  and  against  any  other  form 
or  surface,  and  in  fact  include  everything  that  may 
legitimately  be  termed  Miter  Cutting. 

In  the  problems  of  this  class  two  conditions  exist, 
which  depend  upon  the  nature  of  the  work.  Accord- 
ing to  t\\e  first,  a  simple  elevation  or  plan  of  the  inter- 
secting parts  shows  the  miter  line  in  connection  with 
Uie  profile,  which  is  all  that  is  necessary  to  begin  at 
once  with  the  work  of  laving  out  the  patterns. 

It  frequently  happens,  however,  that  moldings  are 
brought  obliquely  against  sloping  or  curved  surfaces 


in  such  a  manner  that  no  view  can  be  drawn  in  which 
the  miter  line  will  appear  as  a  simple  straight  line. 
Hence  it  becomes  necessary  to  produce  by  the  inter- 
section of  lines  a  correct  elevation  of  the  intersections 
of  the  various  members  of  the  molding,  which  when 
done  results  in  the  much  sought  miter  line.  Or  it  may 
be  necessary  to  develop  a  correct  profile  of  some  oblique 
member  or  molding  in  order  to  effect  a  perfect  miter. 
Thus  some  preliminary  drawing  must  be  done  before 
the  work  of  laying  out  the  miter  patterns  can  be  prop- 
erly begun,  which  constitutes  the  wr</m/  condition  above 
referred  to  and  forms  the  great  reason  whv  the  patten. 
draftsman  should  understand  the  principles  of  projec- 


Pattern  Problems. 


97 


lion,    which   have  been   simplified   for  his    benefit   in 
Chapter  III. 

In  the  arrangement  of  the  problems  those  which 
fulfill  the  first  condition  will  precede  those  of  the 
second,  and  all  of  a  similar  nature  will,  so  far  as  pos- 
sible, be  placed  near  together,  so  that  the  reader, 


knowing  the  kind  that  is  wanted,  will  be  able  to  find 
it  with  little  difficulty.  It  will  also  be  to  his  advantage 
before  reading  any  of  the  problems  in  this  chapter  to 
read  carefully  the  Requirements  and  the  general  Rule 
governing  this  class  of  problems  given  in  Chapter  V 
on  pages  76  and  77. 


PROBLEM  I. 


A  Butt  Miter  Against  a  Plain 

Let  A  B  L  K  in  Fig.  272  be  the  elevation  of  a 
portion  of  a  cornice,  of  which  C  D  is  the  profile  and 
A  B  the  angle  or  inclination  of  the  surface  against 
which  the  cornice  is  required  to  miter.  Divide  the 
curved  parts  of  the  profile  into  spaces  in  the  usual 
manner,  and  from  all  points  in  profile  draw  lines  parallel 
with  A  K,  cutting  the  miter  line  A  B.  On  any  con- 
venient line,  as  E  F,  at  right  angles  to  the  cornice,  lay 
off  a  stretchout  of  the  profile  C  D,  space  by  space  as 
they  occur,  through  the  points  in  which  draw  the 
measuring  lines,  all  as  indicated  by  the  small  figures. 
Placing  the  J-square  at  right  angles  to  the  lines  of  the 
cornice,  or,  what  is  the  same,  parallel  to  the  stretchout 
line,  bring  it  successively  against  the  points  in  the 
miter  line  A  B  and  cut  measuring  lines  of  corresponding- 
number,  as  indicated  by  the  dotted  lines.  A  line  traced 
through  these  points,  as  indicated  by  H  G,  will  be 
the  pattern  required. 


Surface  Oblique  in  Elevation. 

A 


Fig.  272.— A  Butt  Miter  Against,  a  Plain  Surface  Oblique  in  Elevation 


PROBLEM  2. 

A  Butt  Miter  Against  a  Plain  Surface  Oblique  in  Plan. 
R 


ill. 

iliih 


B 


Fig.  171.— A.  Butt  Miter  Against  a  Plain  Surface  Oblique  in  Plan. 


Let  A  B  L  K  in  Fig.  273  be  the 
plan  of  the  cornice  which  is  required 
to  miter  against  a  vertical  surface  stand- 
ing at  any  angle  with  the  lines  of  the 
cornice,  the  angle  being  shown  by  A  B. 
Draw  the  profile  C  D  in  position  corresponding  to  the 
lines  of  the  cornice,  all  as  indicated.  Space  the  profile  in 
the  usual  manner,  and  through  the  points  draw  lines 
parallel  to  the  direction  of  the  cornice,  cutting  the  miter 
line  A  B.  On  any  convenient  line  at  right  angles  to  the 
lines  of  the  cornice  lay  off  the  stretchout  E  F  of  the  pro- 
file C  D,  through  the  points  in  which  draw  measuring 
lines  in  the  usual  manner.  Placing  the  T-square  at  right 
angles  to  the  cornice,  or,  what  is  the  same,  parallel  to 
the  stretchout  line  E  F,  bring  it  successively  against  the 
points  in  A  B  and  cut  the  corresponding  measuring  lines. 
A  line  traced  through  the  points  of  intersection  thus  ob- 
tained, shown  by  H  G,  will  be  the  pattern  required. 


98  The  New  Metal   Worker  Pattern  Book. 

PROBLEM   3. 

A  Square  Return  Miter,  or  a  Miter  at  Right  Angles,  as  in  a  Cornice  at  the  Corner  of  a  Building. 


In  Fig.  274,  let  A  B  D  C  be  the  elevation  of  a 
cornice  at  the  corner  of  the  building  for  which  a  miter 
at  right -angles  is  desired.  As  has  been  explained  in  the 
chapter  on  the  Principles  of  Pattern  Cutting  (page  77), 
the  process  of  cutting  a  miter  for  a  right  angle  admits 
of  certain  abbreviations  not  employed  when  other 


Fig,  274. — A  Square  Return  Miter,  as  in  a  Cornice  at  the 
Corner  of  a  Building. 

angles  are  required.  The  demonstration  here  intro- 
duced is  calculated  to  show  the  method  of  obtaining 
the  pattern  for  a  square  miter  with  the  least  possible 
labor.  Divide  the  profile  A  B  into  any  convenient 
number  of  parts,  as  shown  by  the  small  ligures.  At 
right  angles  to  the  lines  of  the  molding,  and  in  con- 
venient proximity  to  it,  lay  off  the  stretchout  E  F, 


through  the  points  in  which  draw  measuring  lines  in 
the  usual  manner,  parallel  to  the  lines  of  the  cornice, 
producing  them  far  enough  to  intercept  lines  dropped 
vertical!  v  from  points  in  A  B.  Place  the  7-s(lU!llv-  :if 
right  angles  to  the  cornice,  or,  what  is  the  same,  parallel 
to  the  stretchout  line,  and,  bringing  it  successively 
against  all  the  points  in  the  profile  A  B,  cut  measuring 
Hues  of  corresponding  numbers.  Then  a  line  traced 
through  these  points,  as  shown  by  G  .11,  will  be  tin- 
pattern  sought.  The  reason  for  this  is  as  follows:  As 
the  angle  of  this  miter  cannot-  be  shown  in  any  other 
view  than  a  plan,  the  plan  is  the  correct  view  from 
which  to  derive  the  pattern;  having  drawn  which,  as 


Fig.  £75.— Plan  of  a  Square  Return  Miter. 

shown  iu  ,Fig.  275,  the  operation  of  developing  the 
pattern  becomes  exactly  the  same  as  in  the  previous 
problem  (Fig.  273).  In  Fig.  274,  A  B  DC  represents 
the  elevation  of  a  portion  of  a  cornice,  while  A  B  rep- 
resents the  profile  of  the  return  or  receding  portion 
against  which  the  piece  A  B  D  C  is  required  to  miter, 
or,  in  other  words,  the  miter  line.  As  the  profiles  of 
the  face  piece  and  of  the  return  piece  are  of  course  the 
same,  the  outline  A  B  becomes  at  once  the  profile  and 
the  miter  line;  therefore  that  portion  of  the  rule  which 
says,  "  drop  the  points  from  the  profile  on  to  the  miter 
line,"  must  be  omitted.  All  that  remains  then  is  to 
drop  the  points  at  once  into  the  stretchout. 


PROBLEM  4. 

A  Return  Miter  at  Other  Than  a  Right  Angle,  as  in  a  Cornice  at  the  Corner  of  a  Building. 


In  Fig.  276,  let  A  B  C  D  be  the  elevation  of  a 
portion  of  cornice,  and  let  G  II  K  be  the  plan  of  any 
angle  around  which  the  cornice  is  to  be  carried,  a  pattern 
being  required  for  an  arm  of  the  miter.  Complete 


the  plan  by  drawing  the  lines  E  F  and  F  L,  inter- 
secting at  F,  giving  the  correct  projection  of  the  mold- 
ing from  G  H  and  H  K,  and  then  draw  the  miter  line 
between  the  points  H  and  F.  It  will  be  observed  that 


Pattern  Problems. 


99 


the  arm  G  H  F  E  has  been  projected  directly  from  the 


profile  A  B,  thus 
placing  profile  and 
plan  in  correct  rela- 
tion to  each  other. 
Divide  the  profile  A 
B  in  the  usual  man- 
ner into  any  coiiven- 


F 

- 

"" 

V 

~V 

• 

—  — 

-N. 

i 

] 

g 

: 

- 

-     '. 

in 

1! 

12 

13 

14  15 



— 

- 

- 

- 

- 

— 

- 



-_,_ 

x 

7 

:~" 

P- 

- 

x 



;» 

Fig.  276.— A  Return  Miter  at  Other  than  a  Right  Angle,  as 
in  a  Cornice  at  the  Corner  of  a  Building. 


ient  number  of  parts,  and  from  the  points  thus  ob- 
tained drop  lines  vertically  on  to  the  miter  line  in  the 
plan  F  H,  as  shown.  At  right  angles  to  this  arm  of 
the  cornice,  as  shown  in  plan,  lay  off  a  stretchout  of 
the  profile,  as  shown  by  N  M,  through  the  points  in 
which  draw  the  usual  measuring  lines,  as  indicated. 
Place  the  T-square  parallel  to  this  line,  or,  what  is 
the  same,  at  right  angles  to  E  F,  and,  bringing  it  suc- 
cessively against  the  points  in  F  H,  cut 
measuring  lines  of  corresponding  num- 
bers. Then  a  line  traced  through  the 
points  thus  obtained,  as  shown  by  0  P, 
will  be  the  pattern  sought.  As  inti- 
mated at  the  outset  of  this  problem,  the 
angle  G  H  K  represents  any  angle  what- 
ever, and  the  course  to  be  pursued  is 
exactly  the  same  whether  it  be  acute 
or  obtuse.  Of  course  the  more  acute 
the  angle  G  H  K  the  longer  will  the 
miter  line  H  F  become,  as  may  be  ascer- 
tained by  experiment,  producing  a  cor- 
responding increase  in  the  projection  of 
the  different  parts  of  the  pattern  from 
the  line  N"  M. 


PROBLEM   5. 

A  Butt  Miter  Agfainst  a  Curved  Surface. 


In  Fig.  277,  let  A  B  be  the  profile  of  any  cornice 
and  D  K  H  C  be  the  elevation  of  the  same  showing  the 
curved  surface  C  D,  against  which  it  is  required  to 
miter.  The  principle  herein  involved  is  exactly  the 
same  as  that  in  Problem  1.  Space  the  profile  in  the  usual 
manner,  and  through  the  points  draw  lines  cutting  C  D. 
At  right  angles  to  the  line  of  cornice  lay  off  the  stretch- 
out L  M,  as  shown,  through  the  points  in  which 
1 1  raw  measuring  lines  in  the  usual  manner.  Place  the 
J-square  parallel  to  the  stretchout  line,  or,  what  is  the 
same,  at  right  angles  to  the  lines  of  the  cornice,  and, 
bringing  it  against  the  several  points  in  C  D,  cut  the 
corresponding  measuring  lines,  as  shown.  In  the 
event  of  a  wide  space,  as  shown  by  a'  bl  in  the  eleva- 
tion, the  curve  between  these  points  may  be  trans- 
ferred to  the  pattern  by  means  of  a  piece  of  tracing 
paper,  or,  if  it  is  a  regular  curve,  its  radius  may  be 
used  as  shown  by  G  and  G'and  the  arrow  points.  A  line 
traced  through  the  several  points  of  intersection,  as 
shown  by  E  F,  will  be  the  shape  of  the  required  pattern. 


--G' 

F  I,;  H' 

Fig.  X77.—A  Butt  Miter  against  a  Regular  Curved  Surfae*. 


100 


The  New  Metal   Worker  Pattern  Book. 


fig.  tfS.—Thc  Patterns  for  a  Hip  Finish  in  a  Curved  Hansard  Roof,  the  Angle  uf  the  Hip  being  a  Right  Angle. 


Pattern  Problems. 

PROBLEM   6. 


101 


The  Pattern  for  a  Hip  Finish  in  a  Curved  Mansard  Roof,  the  Plan  of  the  Hip  Being  a  Right  Angle. 


The  solution  of  all  problems  concerning  mansard 
roofs,  and  especially  those  in  which  the  roof  surface  is 
curved,  calls  for  much  good  judgment  on  the  part  of 
the  pattern  cutter,  for  the  reason  that  the  original 
designs  that  come  into  his  hands  are  seldom  drawn 
mathematically  correct.  The  upper  part  of  a  mansard 
dome,  such  as  is  shown  in  Fig.  278,  as  it  curves 
away  from  the  eye,  becomes  so  much  flattened  in 
appearance  that,  if  drawn  correctly,  it  might,  to  any 
but  an  expert  draftsman,  create  a  false  impression  of 
the  design  intended ;  hence  the  original  drawing  must 
often  be  taken  for  what  it  means  rather  than  for  what 
it  says. 

The  engraving  represents  an  elevation  of  a  curved 
hip  molding  occurring  in  a  roof,  of  which  E  D  is  the 
vertical  hight  and  M"  K3  is  a  section.  The  first  step 
to  be  described  is  the  method  of  obtaining  the  pattern 
of  the  fascias  of  the  hip  molding.  For  this  purpose  is 
shown  in  the  drawing  such  a  representation  of  it  as 
would  appear  if  the  two  fascias  formed  a  close  joint 
upon  the  angle  of  the  roof,  supposing  that  the  hip 
molding  or  the  bead  is  to  be  added  afterward  on  the 
outside  over  this  joint.  The  part  to  be  dealt  with  may 
be  considered  the  same  as  though  it  were  the  section 
of  a  molding,  instead  of  a  section  of  a  roof,  and  the 
operations  performed  are  identical  with  those  employed 
in  cutting  a  square  miter.  Space  the  profile  II  K  into 
any  convenient  number  of  parts,  introducing  lines  in 
the  upper  part  in  connection  with  the  ornamental 
corner  piece,  shown  by  L  D,  at  such  intervals  as  will 
make  it  possible  to  take  measurements  required  to 
describe  the  shape  of  it  in  the  pattern.  From  this 
profile,  by  means  of  the  points  just  indicated,  lay  off  a 
stretchout,  as  shown  by  H'  K1,  and  through  the  points 
draw  the  usual  measuring  lines.  Bring  the  T-square 
against  the  several  points  in  H  K,  and  cut  the  corre- 
sponding lines  drawn  through  the  stretchout  just 
described.  Then  a  line  traced  through  these  points, 
as  shown  by  H"  K",  will  be  the  outside  line  of  the 
fascia.  For  the  inside  line  take  the  given  width  of  the 
fascia  and  set  it  off  from  this  line  at  intervals,  measur- 
ing at  right  angles  to  it,  as  indicated  by  A1  B',  and 
not  along  the  measuring  lines  of  the  stretchout,  as 
would  be  indicated  by  A'  C.  Then  a  line  traced 
through  these  points,  as  shown  from  M'  to  L',  will  be 


the  inside  line  of  the  fascia  strip.  The  points  in  the 
ornamental  corner  piece  from  L1  to  D1  are  to  be 
obtained  from  the  elevation,  in  case  a  correct  elevation 
is  furnished  the  pattern  cutter,  by  measurement  along 
the  lines  drawn  horizontally  through  the  several  points 
in  L  D,  which  are .  transferred  to  the  measuring 
lines  of  corresponding  number  in  the  stretchout  already 
referred  to.  Or  the  shape  from  L1  to  D1  may  be 
described  arbitrarily  upon  the  pattern  at  this  stage  of 
the  operation,  according  to  the  finish  required  upon 
the  roof.  The  latter  method  is  the  preferable  one.  The 
method  of  constructing  the  elevation,  by  working  back 
from  the  outline  thus  established,  is  clearly  indicated 
by  the  dotted  lines  in  the  engraving.  From  the 
several  points  in  the  profile  H  K  horizontal  lines  are 
drawn,  as  shown,  and  from  the  intersections  of  the 
inside  line  of  the  pattern  of  the  fascia  piece  with  the 
various  measuring  lines,  as  above  described,  lines  are 
dropped,  cutting  these  horizontal  lines  of  correspond- 
ing numbers.  Then  a  line  traced  through  these  points, 
shown  from  M  to  L,  will  be  the  inside  line  of  the 
fascia  piece  in  elevation.  To  cut  the  flange  strip 
bounding  the  fascia  and  corner  piece,  commonly  called 
the  sink  strip,  an  elevation  of  which  is  shown  in  the 
section  from  M"  to  D",  the  following  method  will  be 
the  simplest,  and  at  the  same  time  sufficiently  accurate 
for  all  purposes :  Draw  the  line  G  F  approximately 
parallel  to  the  upper  part  of  the  section  M"  D',  making 
it  indefinite  in  length,  which  cut  by  lines  drawn  from 
the  several  points  in  M'  Da,  at  right  angles  to  it,  as 
shown.  From  F  G,  upon  the  several  lines  drawn  at 
right  angles  to  it,  set  off  spaces  equal  to  the  distance 
upon  lines  of  corresponding  number  from  D  E  to  the 
line  M  L  of  the  elevation.  Then  a  line  traced  through 
these  points,  as  indicated  by  M*  L',  will  constitute  a  pro- 
file of  this  flange  strip.  In  like  manner  set  off  in  contin- 
uation of  it,  the  lengths  measured  from  points  in  the 
ornamental  corner  piece  to  D  E,  all  as  shown  by  L'D1  F. 
From  this  profile  lay  off  a  stretchout  parallel  to  G  F,  as 
shown  by  M*  D',  through  the  points  in  which  draw 
measuring  lines  in  the  usual  manner.  Place  the 
J-square  parallel  to  the  stretchout  line,  and,  bringing 
it  successively  against  points  in  both  the  inner  and  the 
outer  lines  of  the  elevation  of  the  flange  strip,  as 
shown  from  M3  D",  cut  the  measuring  lines  of  corre- 


102 


The  Sew  Metal    Worker   Pattern  Book. 


spending    number.      Then  lines   traced   through  these 
points  of  intersection,  as  shown  from  Ms  to  D",  will  be 


the  pattern  of  the   flange   strip  bounding  the  edge  of 
the  fascia. 


PROBLEM  7. 


Miter  Between  Two  Moldings  of  Different  Profiles. 


To  construct  a  square  miter  between  moldings  of 
dissimilar  profiles  requires  two  distinct  operations. 
The  miter  upon  each  piece  is  to  be  cut  as  it  would  ap- 
pear when  intersected  by  the  other  molding.  Let  A  B 
and  A'  B1  in  Figs.  279  and  280  be  the  profiles  of  two 
moldings,  between  which  a  square  miter  is  required. 
As,  of  course,  the  two  arms  of  the  miter  are  different, 
it  will  be  necessary  to  draw  an  elevation  of  each  show- 
ing the  proper  outline  against  which  it  is  to  miter. 
Beginning,  therefore,  with  the  profile  A  B,  project 
from  it  an  elevation,  as  shown  by  F  C  D  E,  Fig.  279, 
terminating  such  elevation  by  the  profile  of  the  other 
molding,  A1  B',  as  shown  by  F  E.  Then,  as  in  the 
case  of  Problems  1  and  6,  the  line  F  E  becomes  the 


through  these  points,  as  shown  by  F'  E1,  will  give  the 
shape  of  the  cut  on  the  piece  A  B  to  fit  against  the 
profile  E  F.  For  the  other  piece  proceed  in  the  same 
manner,  reversing  the  order  of  the  profiles.  From  its 
profile  A1  B'  produce  the  elevation  K  M  N  L,  Fig.  280, 
completing  same  by  means  of  the  profile  of  the  first 
molding,  A  B,  as  shown  by  M  N.  Divide  A'  B'  m 
the  usual  manner.  Through  the  points  draw  lines  cut- 
ting M  N.  At  right  angles  to  this  piece  lay  off  the 
stretchout  O  P  of  the  profile  A1  B1,  through  the  points 
in  which  draw  measuring  lines,  as  shown.  With  the 
J-s  uare  at  right  angles  to  the  line  K  M  or  N  L,  and 
brought  against  the  points  in  M  N,  cut  corresponding 
measuring  lines  drawn  through  O  P.  A  line  traced 


Fig.  S79. 


Fig.  280. 


Miter  between  Two  Moldings  of  Different  Profiles. 


miter  line,  and  the  method  of  procedure  is  the  same 
as  in  those  problems.  Divide  A  B  into  any  convenient 
number  of  parts  in  the  usual  manner,  from  which  carry 
lines  horizontally  against  F  E.  At  right  angles  to  the 
lines  of  the  molding  lay  off  a  stretchout,  Gr  II,  of  the 
profile  A  B,  through  the  points  in  which  draw  the 
usual  measuring  lines.  Bring  the  T-square  against 
the  points  of  intersection  in  the  line  E  F,  and  cut  the 
corresponding  measuring  lines.  Then  a  line  traced 


through  these  points,  as  shown  by  M'  N',  will  be  the 
shape  of  the  end  of  the  piece  required  to  fit  against  the 
profile  M  N.  In  the  event  of  the  points  obtained  by 
spacing  the  profiles  A  B  and  A1  B1  not  meeting  all  the 
points  in  the  profiles  F  E  and  M  N  necessary  to  be 
marked  in  the  pattern,  then  lines  must  be  drawn  back- 
ward from  such  points  in  profiles  M  N  and  E  F,  cut- 
ting the  profile  A'  B1  or  A  B,  as  the  case  may  be. 
Corresponding  points  are  then  to  be  inserted  in  the 


Pdttmt    Pro/items. 


103 


stretchouts,  through  which  measuring  linos  arc  to  In- 
drawn, which,  in  turn,  ;nv  ID  In-  intersected  1>\-  lines 
dropped  from  the  points.  An  illustration  of  this  occurs  in 
Fig.  280,  where  it  will  be  seen  that  no  point  obtained  by 
the  dividing  of  the  profile  A'  B1  strikes  the  point  X  of 
the  miter  line,  which  is  absolutely  necessary  to  the 
shape  of  the  pattern.  Therefore,  after  spacing  the 
profile,  a  line  is  drawn  from  X  back  to  A1  B',  foring  the 
point  No.  6£.  In  turn  this  point  is  transferred  to  the 


stretchout  O  P,  also  marked  6£,  from  which  a  measur- 
ing line  is  drawn  in  the  same  manner  as  through  the 
other  points  in  the  stretchout,  upon  which  a  point  from 
X  is  dropped,  as  shown  by  X'.  In  actual  practice  such 
expedients  as  this  must  be  resorted  to  in  almost  every 
case,  because  usually  there  is  less  correspondence  be- 
tween the  members  of  dissimilar  profiles,  between  which 
a  miter  is  required,  than  in  the  illustration  here  given. 
By  this  means  profiles,  however  unlike,  can  be  joined. 


PROBLEM  8. 


A  Butt  Miter  Against  an  Irregular  or  Molded  Surface. 


Let  B  A  in  Fig.  281  be  the  profile  of  a  cornice, 
against  which  a  molding  of  the  profile,  shown  by  G  H,  is 
to  miter,  the  latter  meeting  it  at  an  inclination,  as  indi- 


Fig.  £81- — -4  Butt  Miter  arjainst  an  Irregular  or  Molded  Surface. 

cated  by  C  D.  Construct  an  elevation  of  the  oblique 
molding,  as  shown  by  C  D  F  E,  in  line  with  which 
draw  the  profile  G  II.  Divide  &  H  in  the  usual  man- 


ner into  any  convenient  number  of  parts,  and  through 
the  points  draw  lines  parallel  to  the  lines  of  the  in- 
clined molding,  cutting  the  profile  B  A,  all  as  indicated 
by  the  dotted  lines.  At  right  angles  to  the  lines  of  the 
molding,  of  which  a  pattern  is  sought,  lay  off  a  stretch- 
out, M  N,  in  the  usual  manner,  through  the  points  in 
which  draw  measuring  lines.  Place  the  T-square  at 
right  angles  to  the  lines  of  the  inclined  molding,  or, 
what  is  the  same,  parallel  to  the  stretchout  line,  and, 
bringing  it  against  the  points  of  intersection  formed  by 
the  lines  drawn  from  the  profile  G  II  across  the  profile 
B  A,  cut  the  corresponding  measuring  lines.  In  the 
event  of  any  angles  or  points  occurring  in  the  profile 
B  A  which  are  not  met  by  lines  drawn  from  the  points 
in  G  H,  additional  lines  from  these  points  must  be 
drawn,  cutting  the  profile  G  H,  in  order  to  establish 
corresponding  points  in  the  stretchout.  Thus  the 
points  3  and  13  in  the  profile  G  II  arc  inserted  after 
spacing  the  profile,  as  described  in  Problem  7,  because 
the  points  with  which  they  correspond  in  the  profile 
B  E  are  angles  which  must  be  clearly  indicated  in  the 
pattern  to  be  cut.  Having  thus  cut  the  measuring 
lines  corresponding  to  the  points  in  the  profile  B  A, 
draw  a  line  through  the  points  of  intersection,  as  shown 
by  O  P.  Then  O  P  will  be  the  shape  of  the  pattern  of 
the  incline  cornice  to  miter  against  the  profile  A  B. 


104 


77;e  New  Metal    Worker  Pattern  Book. 

PROBLEM  9. 

The  Pattern  of  a  Rectangular  Flaring:  Article. 


In  Fig.  282,  let  C  A  B  E  be  the  side  elevation  of 
the  article,  of  which  F  I  K  M  is  the  plan  at  the  base 
and  G  II  L  N  the  plan  at  the  top.  Let  it  be  required 
to  produce  the  pattern  in  one  piece,  the  top  included. 
Make  H1  L1  N1  G1  in  all  respects  equal  to  II  L  N  G  of 
the  plan.  Through  the  center  of  it  likewise  draw  E  P 
indefinitely,  and  through  the  center  in  the  opposite 
direction  draw  0  S  indefinitely.  From  the  lines  H1  L1 
and  G1  N1  set  off  T  0  and  W  S  respectively,  each  in 
length  equal  to  the  slant  hight  of  the  article,  as  shown 
by  C  A  or  E  B  of  the  elevation.  Through  O  and  S 
respectively  draw  I1  K1  and  F1  M1,  parallel  to  H1  L1  and 
G1  N1,  and  in  length  equal  to  the  corresponding  sides  in 
the  plan  I  K  and  F  M,  placing  one-half  that  length 
each  way  from  the  points  0  and  S.  In  like  manner  set 
off  V  P  and  U  E,  also  equal  to  C  A,  and  draw  through 
E  and  P  the  lines  F"  I'  and  K''  M",  parallel  to  the  ends 
of  the  pattern  of  the  top  part  as  already  drawn,  and  in 
length  equal  to  I  F  and  K  M  of  the  plan.  Draw  I'  H1, 
K1  L1,  K'  L',  M3  N1,  M1  N',  F1  G1,  F1  G1  and  I'  H1, 
thus  completing  the  pattern  sought.  In  the  same 
general  way  the  pattern  may  be  described,  including 
the  bottom  instead  of  the  top,  if  it  be  required  that 
way. 

Considering  this  problem  in  the  light  of  miter  cut- 
ting proper,  I  H  G  F  and  F  G  N  M  may  be  regarded  as 
the  plan  of  two  similar  moldings  of  which  A  C  is  the 
profile,  I  H,  G  F  and  N  M  being  the  miter  lines.  0  T 
is  the  stretchout  line,  drawn  at  right  angles  to  F 
M,  while  I1  K1  and  H1  L'  are  the  measuring  lines 
representing  respectively  the  points  C  and  A  of  the 
profile. 


The  points  F  and  G  are  then  dropped  into  their 
respective  measuring  lines,  thus  locating  the   points  I' 


LLE    AT  ON. 


PATTERN. 


Fig.  S82. — The  Pattern  of  a  Rectangular  Flaring  Article. 

and  H'  at  one  end  of  the  pattern,  while  points  K1  and 
L1  are  derived  from  M  and  N  at  the  other  end. 


PROBLEM  10. 


Patterns  of  the  Face  and  Side  of  a  Plain  Tapering  Keystone. 


Let  A  B  D  C  in  Fig.  283  be  the  elevation  of  the 
face  of  a  keystone,  and  G  E'  F*  K  of  Fig.  284  a  sec- 
tion of  the  same  on  its  center  line. 

Sometimes  problems  occur  which  are  so  simple 
that  it  is  not  apparent  that  their  solution  is  an  ex- 
emplification of  any  rule.  That  this,  with  others  in 
which  plain  surfaces  form  the  largest  factors,  may  be 


so  designated,  will  be  sufficient  excuse  for  a  brief  ref- 
erence to  first  principles.  This  problem  is  generally 
referred  to  as  finding  the  "  true  face  "  of  the  keystone, 
because,  the  face  being  inclined,  the  elevation  ABC 
D  does  not  represent  the  ' '  true  face  "  or  "  true  ' ' 
dimensions  of  the  face.  To  state  the  case,  then,  in  con- 
formity with  the  rule,  A  B  and  C  D  are  the  upper  and 


Pattern  Probkms. 


105 


lower  lines  of  a  molding,  of  which  Ea  F*  of  Fig.  284  is 
the  profile,  and  A  G  and  B  D  are  the  surfaces  against 


Fig.  284. 
Patterns  of  the  Face  and  Side  of  a  Plain  Tapering  Keystone. 

which  it  miters,  or  the  miter  lines.  Therefore,  to  lay 
out  the  pattern,  draw  any  line,  as  E'  F1,  at  right  angles 
to  A  B  for  a  stretchout  line,  upon  which  lay  off  the 
stretchout  taken  from  the  profile  E'  F',  Fig.  284,  which 


in  this  case  consists  of  only  one  space,  as  shown  by 
E'  F' ;  through  the  points  E'  and  F1  draw  the  hori- 
zontal lines  A'  B'  and  C'  D',  which  are  none  other  than 
the  measuring-  lines.  Then,  with  the  T-square  placed 
parallel  with  the  stretchout  line,  drop  the  points  from 
the  miter  lines  A  C  and  B  D  into  lines  of  correspond- 
ing letter,  which  connect,  as  shown  by  A1  C1  and  B1  D1, 
which  completes  the  pattern. 

In  developing  the  pattern  for  the  side,  W  G  and 
F3  K  are  the  lines  of  the  molding,  B  D  of  Fig.  283  its 
profile  and  E"  Fa  the  miter  line.  Hence  upon  any 
vertical  line,  as  L  K',  lay  off  the  stretchout  of  profile 
B  D,  locating  the  points  M'  and  H*,  all  as  shown  by 
L  M  H1  K1,  through  which  points  draw  the  measuring 
lines;  then,  with  the  T-square  placed  parallel  to  L  K', 
drop  the  points  E1  and  FJ  into  lines  of  corresponding 
letter,  as  shown  by  E3  F'.  As  the  vertical  lines  at  Gr- 
and K  represent  the  position  of  surfaces  against  which 
the  side  is  required  to  fit  at  the  back,  bring  the  T-square 
against  each,  thus  locating  them  in  the  pattern  at  G1 
and  K',  as  shown. 

As  the  side  must  also  fit  over  the  molding  of  the  arch 
an  opening  must  be  cut  in  it  corresponding  in  shape 
to  the  profile  of  the  arch  molding  N,  which  is  given  in 
the  sectional  view.  It  is  therefore  only  necessary  to 
transfer  this  profile  to  the  pattern,  placing  the  top  at 
the  measuring  line  M  and  the  bottom  at  the  measuring 
line  H',  all  as  shown  at  N1. 


PROBLEM  II. 

Patterns  for  the  Corner  Piece  of  a  Mansard  Roof,  Embodying:  the  Principles  Upon  Which  All  Mansard 

Finishes  are  Developed. 


One  of  the  first  steps  in  developing  the  patterns 
for  trimming  the  angles  of  a  mansard  roof  is  to  obtain 
a  representation  of  the  true  face  of  the  roof.  In  other 
words,  inasmuch  as  the  surface  of  the  roof  has  a  slant 
equal  to  that  shown  in  the  profile  of  the  return,  the 
length  of  the  hip  is  other  than  is  shown  in  the  eleva- 
tion, and  this  difference  in  dimensions  extends  in  a  pro- 
portionate degree  to  the  lines  of  the  various  parts  form- 
ing the  finish.  Not  only  are  the  vertical  and  oblique 
dimensions  different,  but,  as  the  result  of  this,  the 
angle  at  A  is  different  from  that  shown  in  a  normal  ele- 
vation. Hence,  it  is  of  the  greatest  importance  to  ob- 
tain a  "  true  face  "  or  elevation  of  the  roof  as  it  would 


appear  if  swung  into  a  vertical  position,  which  may  be 
accomplished  as  follows : 

In  Fig.  285,  let  A  E  F  C  be  the  elevation  of  a 
mansard  roof  as  ordinarily  drawn,  and  let  A1  G  be  the 
profile  showing  the  pitch  drawn  in  line  with  the  eleva- 
tion. Set  the  dividers  to  the  length  A1  G,  and  from 
A1  as  center,  strike  the  arc  G  G1,  letting  G'  fall  in  a 
vertical  line  from  A1.  From  G'  draw  a  line  parallel  to 
the  face  of  the  elevation,  as  shown  by  G1  C1,  and  from 
the  several  points  in  the  hip  finish,  as  shown  by 
C  and  K,  drop  lines  vertically,  cutting  G1  C1  in  the 
points  C1  and  K',  as  shown.  From  these  points  carry 
lines  to  corresponding  points  in  the  upper  line  of  the 


106 


The  New  Metal   Worker  Pattern  Book. 


elevation,  as  shown  by  C'  A  and  K'  h.  Then  A  C'  F1 
E  represents  the  pattern  of  the  surface  shown  by  A  C 
F  E  of  the  elevation.  In  cases  where  the  whole  hight 
of  the  roof  cannot  be  put  into  the  drawing  for  use,  as 


B'  draw  the  horizontal  line,  as  shown  by  B1  B3,  and 
from  B  drop  a  vertical  line  cutting  this  line,  as  shown, 
in  the  point  Bs.  By  inspection  of  the  engraving  it  will 
be  seen  that  the  point  B3  falls  in  the  line  A  C'  obtained 


Elevation 
Fig.  $85.— The  Plain  Surfaces  of  a  Mansard  Roof  Devetoped. 


above  described,  the  same  result  may  be  accomplished 
by  assuming  any  point  as  far  from  A  as  the  size  of  the 
drawing  will  permit,  as  B,  and  treating  the  part  between 
A  and  B  as  though  it  were  the  whole.  That  is,  from 
A,  in  a  vertical  line,  set  off  AB1,  equal  to  A  B.  From 


in  the  previous  operation,  thus  demonstrating  that  the 
latter  method  of  obtaining  the  angle  by  which  to  pro- 
portion the  several  parts  results  the  same  as  the  method 
first  described,  and  therefore  may  be  used  when  more 
convenient. 


PROBLEM  12. 


A  Face  Miter  at  Right  Angles,  as  in  the  Molding  Around  a  Panel. 


In  Fig.  286,  let  A  B  D  C  represent  any  panel, 
around  which  a  molding  is  to  be  carried  of  the  profile 
at  E  and  E'.  The  miters  required  in  this  case  are  of 
the  nature  commonly  called  "  face  "  miters,  to  dis- 
tinguish them  from  other  square  miters,  which  can  only 


be  shown  in  a  plan  view.  A  correct  elevation  of  the 
panel  A  B  D  C,  with  the  lines  of  the  molding  carried 
around  the  same,  determines  the  miter  lines  A  F  and 
G  C,  which,  in  connection  with  the  profiles  at  E  and  E' 
are  all  that  is  necessary  to  the  development  of  the  pat- 


Pattern    Problems. 


107 


tern.  The  two  profiles  are  here  drawn,  thus  constitut- 
ing an  entire  section  of  tin-  panel.  Keeause  it  is  usual, 
for  constructive  reasons,  to  cut  the  two  moldings  with 
the  intervening  panel  in  one  piece  where  the  width  of 


V.-M+H 1 

-X4-I-+ 


Fig.  286.— A  Face  Miter  at  Right  Angles,  us  in  the  Molding  Around 

a  Panel. 


the  metal  will  permit  it.  Divide  the  two  profiles  in 
the  usual  manner  into  the  same  number  of  parts,  from 
which  points  draw  lines  parallel  to  the  lines  of  the 
molding,  cutting  the  miter  lines,  as  shown.  For  the 


pattern  o[  tin-  side  corresponding  to  A  B  lay  off  a 
stretchout  at  right  angles  to  it,  as  shown  by  II  K, 
through  which  draw  measuring  lines  in  the  usual  man- 
ner. Place  the  T-square  at  right  angles  to  A  B,  or, 
what  is  the  same,  parallel  to  the  stretchout  line  H  K, 
and,  bringing  it  successively  against  the  several  points 
in  the  miter  line  A  F,  cut  measuring  lines  of  corre- 
sponding number.  Then  a  line  traced  through  these 
points,  as  shown  by  L  M,  will  be  the  pattern  sought. 
The  other  pattern  is  developed  in  like  manner.  It  is 
usual  to  draw  the  stretchout  lines,  K  II  and  K1  H1 
across  the  lines  of  the  moldings  which  they  represent, 
beginning  the  stretchouts  at  the  inner  lines  of  the  mold- 
ing, thus  :  Point  10  of  profile  E  would  be  located  at  V, 
while  point  10  of  profde  E1  would  be  at  W.  While  this 
is  apt  to  produce  some  ^confusion  of  lines  in  actual 
practice,  it  gives  the  entire  profile  in  one  continuous 
stretchout  for  the  purpose  alluded  to  above — that  of 
cutting  the  entire  width  of  the  panel  in  one  piece. 
Should  it  be  desired  to  make  one  of  the  moldings 
separate  from  the  rest,  an  additional  point  for  the  pur- 
pose of  a  lap  is  assumed  at  one  of  the  moldings,  as  11 
of  profile  E'.  The  pattern  for  the  end  piece,  A  C, 
may  be  derived  without  drawing  an  additional  profile, 
as  its  profile  and  stretchout  are  necessarily  the  same  as 
that  of  the  other  two  arms ;  therefore  reproduce  H  K 
on  a  line  at  right  angles  to  A  C,  as  shown  by  N  O, 
through  the  points  in  which  draw  measuring  lines  in 
the  usual  manner,  producing  them  sufficiently  far  in 
each  direction  to  intercept  lines  dropped  from  the 
points  in  the  two  miter  lines.  Place  the  "["-square  at 
right  angles  to  A  C,  and,  bringing  it  successively 
against  points  already  in  A  F  and  C  G,  cut  measuring 
lines  of  corresponding  numbers.  Then  lines  traced 
through  the  intersections  thus  formed,  as  shown  by 
P  R  and  S  T,  will  be  the  shape  of  the  pattern  of  the 
end  piece. 

It  may  be  noticed  in  the  last  operation  that  drop- 
ping the  points  from  either  of  the  miter  lines,  as  A  F, 
into  the  measuring  lines  is,  in  fact,  only  continuing  in 
the  same  direction  the  lines  previously  drawn  from  the 
profile  E  to  the  line  A  F;  and  that  in  reality  the  shape 
of  the  cut  at  P  R  is  developed  without  the  assistance 
of  the  miter  line,  thus  giving  another  instance  of  the 
fact  that  any  square  miter  can  be  cut  by  the  short 
method  when  the  relation  of  the  parts  is  understood. 


108 


The  New  Metal   Worker  Pattern  Book. 

PROBLEM  13. 

The  Patterns  of  the  Moldings  Bounding;  a  Panel  Triangular  in  Shape. 


In  Fig.  287,  let  D  E  F  be  the  elevation  of  a 
triangular  panel  or  other  article,  surrounding  which  is  a 
molding  of  the  profile,  shown  at  G  andG1.  Construct 
an  elevation  of  the  panel  molds,  as  shown  by  ABC, 
and  draw  the  miter  lines  A  D,  B  E  and  C  F.  For  the 
patterns  of  the  several  sides  proceed  as  follows  :  Draw 
a  profile,  Or,  placing  it  in  correct  relative  position  to 
the  side  D  F,  as  shown.  Divide  it  into  any  con- 


of  the  three  sides,  at  convenient  points,  draw  stretch- 
out lines,  as  shown  by  H  I,  H1  I1  and  IP  P,  through 
the  points  in  which  draw  the  usual  measuring  lines. 
With  the  T-square  parallel  to  each  of  the  several 
stretchout  lines,  or,  what  is  the  same,  at  right  angles 
to  the  respective  sides,  bringing  the  blade  successively 
against  the  points  in  the  several  miter  lines,  cut  the 
corresponding  measuring  lines,  all  as  indicated  by  the 


Fig.  tS7.—The  Patterns  of  the  Moldings  Bounding  a  Triangular  Panel. 


venient  number  of  parts  in  the  usual  manner,  and 
through  these  points  draw  lines,  as  shown,  cutting  the 
miter  lines  F  C  and  A  D.  In  like  manner  place  the 
profile  G'  in  a  corresponding  position  relative  to  the 
side  E  F.  Divide  it  into  the  same  number  of  parts, 
and  draw  lines  intersecting  those  drawn  from  the  first 
profile  in  the  line  F  C,  also  cutting  the  line  E  B. 
By  this  operation  points  are  obtained  in  the  three 
miter  lines  A  D,  E  B,  F  C,  from  which  to  lay  off  the 
patterns  in  the  usual  manner.  At  right  angles  to  each 


dotted  lines.  Then  lines  traced  through  the  points  of 
intersection  thus  obtained  will  describe  the  patterns 
required.  A'  C'  F'  D1  will  be  the  pattern  for  the  side 
A  D  F  C  of  the  elevation,  and  likewise  C'  Ba  E' 
F3  is  the  pattern  for  the  side  described  by  similar 
letters. 

Placing  another  profile  in  the  molding  A  B  D  E 
would,  if  divided  the  same  as  the  others,  only  result 
in  another  set  of  intersections  at  the  points  already 
existing  on- the  lines  A  D  and  B  E,  as  occurred  on 


Pattern   Problems. 


109 


the  line  F  C,   hence  to  save  labor  one  profile  in  this 
case  is  all  that  is  really  necessary,  the  points  being 


carried  around  from   that  and  dropped  into  the  three 
stretchouts  respectively. 


PROBLEM   14. 

The  Patterns  of  a  Molding  Mitering  Around  an  Irregular  Four-Sided  Figure. 


In  Fig.  288,  let  A  B  C  D  be  the  elevation  of  an  ir- 
regular four-sided  figure,  to  which  a  molding  is  to  be 
fitted  of  the  profile  shown  by  K.  Place  a  duplicate 
profile  against  the  side  opposite,  as  shown,  from  which 


in  these  several  stretchouts  draw  measuring  lines  in  the 
usual  manner,  producing  them  until  they  are  equal  in 
length  to  the  respective  sides,  the  pattern  of  which  is 
to  be  cut.  Placing  the  T-square  at  right  angles  to  the 


Fig.  tSS.—The  Patterns  of  a  Molding  Mitering  Around  an  Irregular  Four-Sided  Figure. 


project  the  lines  necessary  to  complete  the  elevation 
of  the  molding  as  it  would  appear  when  finished,  all  as 
shown  by  B  F  G  H.  Draw  the  several  miter  lines 
B  F  C  G,  D  H  and  A  E.  Divide  the  two  profiles  into 
the  same  number  of  parts  in  the  usual  njanner,  through 
the  points  in  which  draw  lines  parallel  to  the  lines  of 
the  molding  in  which  they  occur,  cutting  the  miter 
lines,  as  shown.  At  right  angles  to  each  of  the  several 
sides  lav  off  a  stretchout  from  the  profile,  as  shown  by 
L  M,  L'  M',  L'  M',  L3  M3.  Through  the  several  points 


lines  of  the  several  sides,  or,  what  is  the  same,  parallel 
to  the  stretchout  lines,  bring  it  against  the  points  in 
the  miter  lines,  cutting  the  corresponding  measuring 
lines,  all  as  indicated  by  the  dotted  lines.  Then  the 
lines  traced  through  these  points  of  intersection  will 
give  the  several  patterns  required.  Thus  E1  II'  D'  A1 
will  be  the  pattern  of  the  side  E  H  I)  A  of  the  eleva- 
tion; H3  D"  C1  G'  will  be  the  pattern  ot  the  side 
H  D  C  G ;  Ga  F1  B'  C'  that  of  F  B  C  G ;  and  F'  BJ 
A'  E3  that  of  the  remaining  side. 


110 


The  Xcir 


\\'<>rkur  Pattern  Book. 


PROBLEM  15. 


The  Patterns  of  Simple  Gable  Miters. 


In  Fig.  289,  let 'A  B  K  and  15  K  B  be  the  angles 
of  the  miters  at  the  foot  and  peak  of   a  gable.      Draw 
profiles  of  the  required  molding  in  correct  relation  to 
both   the   horizontal  and  inclined  moldings,  as  shown 
at    H    and    H',    through    the    angles    of    which    draw 
the  other  parallel   lines   necessary  to  complete  the  ele- 
vation.    Their  intersection  at  the  base  of  the  gable 
produces  the  miter  line   B    C,    while   the  miter  line  at 
the  top  of  the  gable  is  a  vertical  line,  because  the  two 
sides   of  the  gable,   K    B   and   K   R,    are  of  the  same 
pitch.      The  profile    II    is   so  placed    as   also    to    repre- 
sent   the    return    at    the    side  at  its    proper   distance 
from  B.      Divide    the    profile  11    in    the 
usual   manner  into  any  convenient  nuin-      A 
ber  of  equal  parts.      Place  the  T-square 
parellel   to    the    lines   in   the    horizontal 
molding,    and,    bringing   it  successively 
against  the  points  in  the  profile,  cut  the 
miter   line    B   C,    as    shown.      At    right 
angles    to    the    lines    of    the    hori/.oiital 
cornice  draw  the  stretchout  E  K,  through 
the    points    in    which    draw    the    usual 
measuring    lines,    as    shown.       Reveoe 
the  ^-square,  letting  the  blade  lie  parallel 
to  the  stretchout  line  E  F,  and,  bringing 
it  against  tile  several  points  of  the  profile 
H,    cut    the    corresponding    measuring 
lines.      Then  a  line  traced  through  these 
points  of  intersection,  as  shown  from  G 
to  V,  will  be  the    pattern  of  the  end  of 
the    horizontal     cornice    inhering    with 
the   return.      In    like  manner,    with  the 
T-square   in    the  same  position,  bring  it 
against  the  points  in  the  miter  line  B  C, 
and  cut  corresponding  measuring  lines   drawn  tli rough 
the  same  stretchout.      Then   a  line  traced  through  the 
points  of  intersection  thus  obtained,  as  shown  by  T  U, 
will  be  the  pattern  of  the  end  of  the  horizontal  cornice 
inhering  against  the  inclined  cornice.     Divide  the  pro- 
file H'  into  any  convenient  number  of  equal  parts,  all  as 
indicated  by  the  small  figures.      Through  these  points 
draw  lines  cutting   the   miter  line   B   C,    and  also   the 
miter  line    K   L   at    the    top.      At   right  angles  to  the 
lines  of  the  raking  cornice  place  a  stretchout,  E'  F1,  of 


the  profile  H1,  through  the  points  in  which  draw  the 
usual  measuring  lines,  as  shown.  Place  the  f-dqu&ro 
parallel  to  this  stretchout  line,  and,  bringing  it  suc- 


i 144.4 

1 


Fig.  289.— The  Patterns  of  Simple  Gable  Miters. 

cessively  against  the  points  in  B  C  and  K  L.  cut  the 
corresponding  measuring  lines,  all  as  indicated  by  the 
dotted  lines.  Through  the  points  thus  obtained  trace 
lines,  as  indicated  by  M  N  and  O  P.  Then  M  N  will 
be  the  pattern  for  the  bottom  of  the  raking  cornice 
inhering  against  the  horizontal,  and  O  P  will  be  the 
pattern  for  the  top  of  the  same.  The  pattern  shown 
at  G  V  will  also  be  the  pattern  for  the  return  miter- 
ing  with  A  D  of  the  elevation,  it  being  necessary  only 
to  reverse  it  and  to  .establish  its  length. 


Pattern  Problems. 

PROBLEM  16. 


ill 


The  Pattern  for  a  Pedestal  of  Which  the  Plan  is  an  Equilateral  Triangle. 


should  be  drawn  so  as  to  show  one  side  in  profile  and 
the  plan  placed  to  correspond  with  it.  Draw  the  miter 
lines  E  0  and  G  0.  Divide  the  profile  B  D  into  spaces 
of  convenient  size  in  the  usual  manner,  and  number 
them  as  shown  in  the  diagram.  From  the  points  thus 
obtained  drop  lines,  cutting  E  0  and  G  O.  as  shown. 
Lay  off  the  stretchout  N  P  at  right  angles  to  the  side 
E  G,  and  through  the  points  in  it  draw  measuring  lines. 
Place  the  T-square  at  right  angles  to  E  G,  and,  bring- 
ing it  successively  against  the  points  in  the  miter  lines 
E  0  and  G  0,  cut  the  corresponding  measuring  lines. 
A  line  traced  through  these  points  will  be  the  pattern, 
as  shown  by  H  L  M  K. 

The  principle  involved  in  this  and  several  follow- 
ing problems  is  exactlv  the  same  as  that  of  the  preced- 
ing regular  and  irregular  shaped  panels.  In  this  case 
the  shape  of  the  article  is  shown  in  plan  instead  of  ele- 


Fig.  290.— The  Pattern  for  a  Pedestal  of  which  the  Plan  is  an.  Equilateral  Triangle. 


Let  A  B  D  C  in  Fig.  290  be  the  elevation  of  a 
pedestal  or  other  article  of  which  the  plan  is  an  equi- 
lateral triangle,  as  shown  by  F  E  G.  This  elevation 


vation,  and  the  profile  is  too  large  to  permit  of  its 
being  drawn  within  the  plan,  as  were  the  profiles  of  the 
panel  moldings  in  their  elevations. 


112 


The  Xew  Metal   Worker  Pattern  Book. 


PROBLEM  17. 

The  Pattern  for  a  Pedestal  Square  in  Plan. 


In  Fig.  291,  let  A  B  D  C  be  the  elevation  of  a 
pedestal  the  four  sides  of  which  are  alike,  being  in 
plan  as  shown  by  E  H  G  F,  Fig.  292.  Since  the  plan 
is  a  rectangular  figure  the  miters  involved  are  square 
miters,  or  miters  forming  a  joint  at  90  degrees.  A 
square  miter  admits  of  certain  abbreviations,  the  rea- 
sons for  which  are  explained  in  Problem  3,  as  well  as 
in  Chapter  V,  under  the  head  of  Parallel  Forms.  The 
abbreviated  method  which  is  here  illustrated  is  always 
used.  The  plan  is  introduced  only  to  show  the  shape 
of  the  article,  and  is  not  employed  directly  in  cutting 


Fig.  S92.—The  Plan  of  Square  Pedestal. 

the  pattern.  Space  the  profiles,  shown  in  the  eleva- 
tion by  A  C  and  B  D,  in  the  usual  manner,  numbering 
the  points  as  shown.  Set  off  a  stretchout  line,  L  R,  at 
right  angles  to  the  base  line  C  D  of  the  pedestal, 
through  the  points  in  which  draw  measuring  lines. 
Place  the  T-square  parallel  to  the  stretchout  line,  and, 
bringing  it  successively  against  the  points  in  the  two 
profiles,  cut  the  corresponding  lines  drawn  through  the 
stretchout.  A  line  traced  through  these  points,  as 
shown  by  L  M  0  N  K,  will  be  the  pattern  of  a  side. 


Fiy.  291.—  The  Pattern  for  a  Pedestal,  Square  in  Plan. 


PROBLEM  18. 

The  Patterns  for  a  Vase,  the  Plan  of  Which  is  a  Pentagon. 


In  Fig.  293,  let  S  C  K  T  be  the  elevation  of  a 
vase,  the  plan  of  which  is  a  pentagon,  as  shown  O  C' 
Ca  B  P.  The  elevation  must  be  drawn  in  such  a  man- 


ner that  one  of  the  sides  will    be  shown  in 

Draw  the  plan    in  line  and  in  correspondence  with  it. 

Divide  the  profile  into  spaces  of  convenient  size  in  the 


Pattern   Problems. 


113 


usual  manner  and  number  them.  Draw  the  miter 
lines  C'  II1  and  C*  IF  in  the  plan,  and,  bringing  the 
T-square  successively  against  the  points  in  the  profile, 
drop  lines  across  these  miter  lines,  as  shown  by  the 
dotted  lines  in  the  engraving.  Lay  off  the  stretchout 
M  1ST  at  right  angles  to  the  piece  in  the  plan  which 


the  case  of  a  complicated  profile,  or  one  of  many  dif- 
ferent members,  to  drop  all  the  points  across  one  sec- 
tion of  the  plan  C'  H'  IF  Ca  would  result  in  confusion. 
Therefore  it  is  customary,  in  practice,  to  treat  the 
pattern  in  sections,  describing  each  of  the  several 
pieces  of  which  it  is  composed  independently  of  the 


Fig.  294.— Pattern  for  the  Base. 


Fig.  293.— Pattern  for  the  Upper  Part. 
The  Patterns  for  a  Vase,  the  Plan  of  which  is  a  Pentagon. 


corresponds  to  the  side  shown  in  profile  in  the  eleva- 
tion. Through  the  points  in  it  draw  the  usual  measur- 
ing lines.  Place  the  T-square  parallel  to  the  stretch- 
out line,  and,  bringing  it  against  the  several  points  in 
the  miter  lines  which  were  dropped  from  the  elevation 
upon  them,  cut  the  corresponding  measuring  lines 
drawn  through  the  stretchout.  A  line  traced  through 
the  points  thus  obtained  will  describe  the  pattern.  In 


others.  In  the  illustration  given  the  pattern  has  been 
divided  at  the  point  II,  the  upper  portion  being 
developed  from  the  profile  and  plan,  as  above,  while 
the  lower  part  is  redrawn  in  connection  with  a  section 
of  the  plan,  as  shown  in  Kig.  294.  Corresponding 
letters  in  each  of  the  views  represent  the  same  parts, 
so  that  the  reader  will  have  no  trouble  in  perceiving 
just  what  has  been  done.  Instead  oi  redrawing  a  pov- 


114 


Worker   Pattern 


tion  of  the  elevation  and  plan,  as  has  been  done  in  this 
case,  sometimes  it  is  considered  best  to  work  from  one 
profile  rather  than  to  redraw  a  portion  of  it,  as  tha1 
always  results  in  more  or  less  inaccuracy.  Therefore, 
after  using  the  plan  and  describing  a  part  of  the 
pattern,  as  shown  in  the  operation  explained  above. 


a  piece  of  clean  paper  is  pinned  on  the  board,  cover- 
ing this  plan  and  pattern,  upon  which  a  duplicate 
plan  is  drawn,  from  which  the  second  section  of 
the  pattern  is  obtained.  (J  reat  care,  however,  is  neces- 
sary in  redrawing  portions  of  the  plan  to  insure 
accuracy. 


PROBLEM  19. 


The  Pattern  for  a  Pedestal,  the  Plan  of  Which  is  a  Hexagon. 


sides,  drawn  so  that  one  of  the  sides  will  be  shown  in 
profile.  Place  the  plan  below  it  and  corresponding 
with  it.  Divide  the  profile  shown  in  the  elevation 
into  any  convenient  number  of  spaces  in  the  usual 
manner,  and,  to  facilitate  reference  to  them,  number 
them  as  shown.  Bring  the  T-square  against  the  points 
in  the  profile  and  drop  lines  across  one  section  of  the 
plan,  as  shown  bv  II  X  M.  At  right  angles  to  this 
section  of  the  plan  layoff  the  stretchout  line  .N  <  >. 
through  the  points  in  which  draw  the  usual  meas- 
uring lines.  Place  the  T'scluare  parallel  to  the 
stretchout  line,  and,  bringing  it  successively  against 
the  points  in  the  miter  lines  II  X  and  M  X,  cut  the 
corresponding  measuring  lines,  as  indicated  by  the 
dotted  lines.  Then  a  line  traced  through  the  points 


Fig.  205. — The  Pattern  for  a  Pedestal,  the  Plan  of  which  is  a  Hexagon. 


In   Fig.  295,  let  C  D  F  E  be  the  elevation   of  a      thus  obtained  will  be  the  required  pattern,  as  shown  by 


pedestal  which  it  is  desired  to  construct  of  six  equal 


P  S  T  E. 


Pattern 


115 


PROBLEM    20. 

The  Pattern  for  a  Vase,  the  Plan  of  Which  is  a  Heptagon. 


sides  will  be  shown  in  profile.  In  line  with  it  draw  the 
plan,  placing  it  so  that  it  shall  correspond  with  the 
elevation.  Spaee  the  profile  L  P  in  the  usual  manner, 
and  from  the  points  in  it  drop  lines  crossing  one  sec- 
tion of  ihe  plan,  cutting  the  miter  lines  It  S  and  II  V, 
as  shown.  Lay  oil'  a,  stretchout,  A  H,  at  riirht  anules 
to  the  side  of  the  plan  corresponding  to  the  side  of  the 
vase  shown  in  profile  in  the  elevation.  Through  the, 
points  in  it  draw  tin;  usual  measuring  lines.  Place  the 
"|"-s(|iiare  parallel  to  tins  stretchout  line,  and,  bringing 
it  successive] v  against  the  points  in  the  miter  lines, 
cut  the  corresponding  measuring  lines,  as  shown.  A 
line  traced  through  these  points,  as  shown  by  K  O 


Fig.  296.— The  Pattern  fur  a  Vase,  the  Plan  nf  which  is  a  Heptagon. 

In  Fig.  296,  let  E  L  P  G  be  the  elevation  of  the      W  U,  will   be  the  pattern  of  one  of  the  sides  of  the 
vase,    constructed    in    sneh   a   manner  that  one  of  its   ;  vase. 

PROBLEM   21. 

The  Patterns  for  an  Octagonal  Pedestal. 


Let  K  II  G  W  L  in  Fig.  21)7  be  the  elevation  of 
a  pedestal  octagon  in  plan,  of  which  the  pattern  of  a 
section  is  required.  This  elevation  should  be  drawn 
in  such  a  manner  that  one  side  of  it  will  appear  iu 
profile.  Place  the  plan  so  as  to  correspond  in  all 
respects  with  it.  Divide  the  profile  G  W,  from  which 
the  plan  of  the  side  desired  is  projected,  in  the  usual 
manner,  and  from  the  points  in  it  drop  points  upon 
each  of  the  miter  lines  F  T.  and  P  U  in  the  plan.  Lay 
off  a  stretchout,  B  E,  at  right  angles  to  the  side  of  the 
plan  corresponding  to  the  side  of  the  article  shown  in 
profile  in  the  elevation,  and  through  the  points  in  it 


draw  the  usual  measuring  lines.  Place  the  T-square 
parallel  to  the  stretchout  line,  and,  bringing  it  -suc- 
cessively against  the  points  dropped  upon  the  miter 
lines  from  the  elevation,  cut  the  corresponding  measur- 
ing lines.  A  line  traced  through  the  points  thus 
obtained  will  describe  the  pattern  of  one  of  the  sides 
of  which  the  article  is  composed.  In  eases  where  the 
profile  is  complicated,  consisting  of  many  members, 
and  where  it  is  verv  long,  confusion  will  arise  if  all 
the  points  are  dropped  across  one  section  of  the  plan, 
as  above  described.  It  is  also  quite  desirable  in  many 
cases  to  construct  the  pattern  in  several  pieces.  In 


116 


The  New  Mdal   Worker  Pattern  Book, 


such  cases  methods  which  are  described  in  connection  the  pattern  is  cut 'by  means  of  a  part  of  the  plan 
with  Problem  18  may  be  used  with  advantage.  !  redrawn  above  the  elevation,  thus  allowing  the  use 
In  the  present  case  the  pattern  is  constructed  of  two  of  the  same  profile  for  both.  The  same  letters  refer 


t 

Fig.  297.— The  Patterns  for  an  Octagonal  Pedestal. 


pieces,  being  divided  at  the  point  8  of  the  profile. 
The  lower  part  of  the  pattern  is  cut  from  the  plan 
drawn  below  the  elevation,  while  the  upper  part  of 


to  similar  parts,  so  that  the  reader  will  have  no  diffi- 
culty in  tracing  out  the  relationship  between  the  dif- 
ferent views. 


l'nttt'1-n  PnJ'li  nis. 

PROBLEM    22. 

The  Patterns  for  a  Newel  Post,  the  Plan  of  Which  is  a  Decagon. 


it; 


In  Fig.  298,  let  \V  US  I1  <>  H  T  V  be  the  eleva- 
tion of  a  newel  post  which  is  required  to  be  constructed 
in  ten  parts.  Draw  the  plan  below  the  elevation,  as 
shown.  The  elevation  must  show  one  of  the  sections 


w 


the  plan,  as  shown  by  <i  X  II,  and  cutting  the  two 
miter  lines  (i  X  and  II  X.  Lay  oil'  tin-  stretchout  line 
C  D  at  right  angles  to  G  II,  and  through  it  draw  the 
customary  measuring  lines.  Place  the  T-square  parallel 
to  the  stretchout,  and,  bringing  it.  against  the  several 
points  in  the  miter  lines  G  X  and  H  X,  cut  the  corre- 
sponding measuring  lines.  A  line  traced  through  the 
points  thus  obtained  will  describe  the  pattern.  In  order 
to  avoid  confusion  of  lines,  which  would  result  from  drop- 
ping points  from  the  entire  profile  across  onesection  of 
the  plan,  a  duplicate  of  the  cap  A1  W  is  drawn  in  Fig. 
299  in  connection  with  a  section  of  the  plan,  as  shown 


•    Fiy.  299.— Pattern  of  Cap. 


Plan 


Fig.  298.— The  Patterns  for  a  Newel  Post,  the  Flan  of  which  is  a  Decagon. 


or  sides  in  profile,  and  the  plan  must  be  placed  to  cor- 
respond with  the  elevation.  Space  the  molded  parts  of 
the  profile  in  the  usual  manner,  and  from  the  points  in 
them  drop  lines  crossing  the  corresponding  section  of 


by  G1  X'  H1,  which  are  employed  in  precisely  the  same 
manner  as  above  described,  thus  completing  the  pattern 
in  two  pieces,  the  joint  being  formed  at  the  point  num- 
bered 11  of  the  profile  and  the  stretchout. 


118 


PROBLEM    23. 

The  Patterns  for  an  Urn,  the  Plan   of  Which  is  a  Dodecagon. 


In  Fig.  300,  let  X  A  <!  II  l>e  the  elevation  of  an 
urn  to  1)0  constructed  in  twelve  pieces.  The  elevation 
must  l>e  drawn  so  as  to  sliow  one  side  in  profile.  Con- 
struct the  plan,  as  shown,  1o  correspond  with  it  and 


the  several  points  in  the  miter  lines  X  X  and  OX. 
cut  the  corresponding  measuring  lines.  A  line 
traced  through  the  points  thus  obtained  will  describe 

ihe  pattern  sought.  In  this  illustration  is  shown  a 
method  sometimes  resorted  to  by  pattern  cutters  to 
avoid  the  confusion  resulting  from  dropping  all  tin- 
points  across  one  section  of  the  plan.  The  points  from 
1:5  to  20  inclusive  arc  dropped  upon  the  line  OX. 
The  stretchout  C  D  is  drawn  in  exactly  the  middle  of 
the  pattern — that  is,  it  is  drawn  from  X,  the  central 
point  of  the  plan.  Points  are  transferred  liv  the  T- 
s<{uare  from  ()  X  to  the  measuring  lines  on  one  side  of 


V       Plan 
Fig.  SOO.—T'he  I'ritlrrns  for  an  Urn,  the  Plan  of  wltii-h  is  a  Dodecagon. 


draw  the  miter  lines.  ]>ivide  the  prolile  A  S  (!  into 
spaces  in  the  usual  manner,  and  from  the  points  thus 
obtained  drop  lines  across  one  section,  X  X  0,  of  the 
plan.  Lav  off  the  stretchout  C  D  at  right  angles  to  the 
side  N  O  of  the  plan.  Place  the  T-square  parallel  to 
the  stretchout,  and,  bringing  it  successively  against 


the  stretchout,  the  points  on  the  other  side  being  ob- 
tained by  duplicating  distances  from  C  Don  the  several 
lines.  The  points  1  to  13  are  dropped  on  N  X  only. 
The  stretchout  E  F  is  laid  off  at  right  angles  to  the 
side  MNfrom  the  point  X,  and,  the  J-^piare  being  set 
parallel  to  E  F,  the  points  are  transferred  to  the 


119 


measuring  lines  on  one  side  of  E  F,  while  the  distances      scribed  in  the  lirst  instance.   This  plan  will  be  found  ad- 
on  the  opposite  side  are  set  off  by  measurement,  as  de-        vaiitagcons  in  complicated  and  very   extended  profiles. 


PROBLEM   24. 
The  Pattern  for  a  Drop  Upon  the  Face  of  a  Bracket. 


In  Figs.  I'.ol  and  '.}o-2,  methods  of  obtaining  tlie  II  K.  as  shown  bv  O  P,  and  on  0  P  lay  off  a  stretch- 
return  strip  lifting  around  a.  drop  and  mitering  against  out,  through  the  points  in  which  draw  the  usual 
tin1  face  of  a  bracket  an,'  shown.  •  Similar  letters  in  measuring  lines.  From  the  points  in  the  profile  F  G 
the  two  liguivs  represent  similar  parts,  and  the  follow-  I  carry  lines  in  the  direction  of  the  molding — that  is, 


The  Pattern  for  a  Drop  upon  the  face  of  a  hracket. 


ing  demonstration  may  be  considered  as  applying  to 
both.  Let  A  15  I)  C  be  the  elevation  of  a  part  of  the 
face  of  the  bracket,  and  II  Iv  L  a  portion  of  the  side, 
showing  the  connection  between  the  side  strip  of  the 
dr.>p  E  F  (1  aiul  the  face  of  the  bracket.  To  state  the 
case  simply,  F  G  is  the  profile  and  1ST  M  the  miter 
line,  because  N  M  is  the  outline  of  the  surface  against 
which  the  side  strip  miters.  Then,  following  the  rule, 
divide  F  G  into  any  convenient  number  of  parts  in  the 
usual  manner,  as  shown  by  the  small  figures.  Produce 


parallel  to  K  M — intersecting  the  face  of  the  bracket 
N  M.  .Reverse  the  T-square,  placing  the  blade  parallel 
to  the  stretchout  line  O  P,  and,  bringing  it  suc- 
cessively against  the  points  in  N  M,  cut  the  cor- 
responding measuring  lines,  as  indicated  bv  the 
dotted  lines.  Then  a  line  traced  through  these  several 
points  of  intersection,  as  shown  by  O  R  P,  will 
be  the  pattern  of  the  strip  fitting  around  E  F  G 
and  mitering  against  the  irregular  surface  N  M  of 
the  bracket  face. 


120 


The  New  Metal    Worker  Pattern  Book. 


PROBLEM   25. 
The  Pattern  of  a  Boss  Fitting:  Over  a  Miter  in  a  Molding". 


Let  A  B  C  in  Fig.  303  be  the  part  elevation  of  a 
pediment,  as  in  a  cornice  or  window  cap,  over  the 
miter  and  against  the  molding  and  fascia  in  which  a 
boss,  F  K  G  H,  is  required  to  be  fitted,  all  as  shown 
by  N  0  E  D  of  the  side  view. 

The  outline  F  K  G  H  of  the  boss  is  to  be  con- 
sidered as  the  profile  of  a  molding  running  in  the  direc- 
tion shown  by  D  E  in  the  side  view,  and  mitering 
against  the  surface  of  the  cornice  shown  by  N  0  E. 
For  the  patterns  proceed  as  follows:  Divide  so  much 
of  the  profile  of  the  boss  K  F  H  G  as  comes  against 
the  cornice,  shown  from  K  to  F,  into  any  convenient 
number  of  parts,  and  from  these  points  draw  lines 
parallel  to  D  E — that  is,  to  the  direction  of  the  mold- 
ing under  consideration — until  they  intersect  the 
miter  line  N  O  E,  which  in  this  case  is  the  profile  of 
the  cornice  molding.  As  the  boss  is  so  placed  over 
the  angle  in  the  cornice  molding  that  the  distance  from 
K  to  F  is  the  same  as  that  from  K  to  G,  the  part  of 
the  boss  K  G  will  be  an  exact  duplicate  of  K  F  and 
may  be  duplicated  from  the  pattern  of  K  F  without 
another  side  view  drawn  especially  for  it,  which  would 
have  to  be  done  if  the  boss  was  otherwise  placed. 
Therefore,  extend  the  line  N  D  upon  which  to  lay  off 
a  stretchout  of  K  F  H  G,  dividing  the  portion  K1  F1 
into  the  spaces  shown  at  K  F  of  the.  profile,  through 
which  draw  the  usual  measuring  lines.  Make  the  por- 
tion F1  G1  equal  in  length  to  the  part  FUG  and, 
lastly,  the  portion  G1  K"  a  duplicate  of  F1  K'  reversed, 
as  shown.  Place  the  T-square  parallel  to  the  stretch- 


out line  K'    K",    and,    bringing  it  against   the   several 
points  in  N  0,  cut   corresponding  measuring  lines,   as 


Pig.  SOS.— The  Pattern  of  a  Boss  Fitting  Over  a  Miter  in  a  Molding. 

shown.  Then  lines  traced  through  these  points  of  in- 
tersection, as  shown  by  K1  L  M  Ks,  will  be  the  re- 
quired pattern. 


PROBLEM   26. 

The  Patterns  for  a  Keystone  Having:  a  Molded  Face  With  Sink. 


In  Fig.  304,  let  E  A  B  F  be  the  front  elevation 
of  a  keystone,  as  for  a  window  cap,  of  which  K  L  M  P 
S  R  is  a  sectional  view,  giving  the  profile  of  the  mold- 
ing M  N  O  P,  over  which  it  is  required  to  fit.  The 
sink  in  the  face  extends  throughout  its  entire  length, 
and  is  shown  by  G  H  D  C,  its  depth  being  shown  by 
the  line  K  T  of  the  section.  E  F  H  G  and  A  B  D  C 
thus  become  moldings,  of  which  E  A  and  F  B  are  the 
parallel  lines,  E  F,  G  H,  C  D  and  A  B  the  miter  lines, 
and  K  R  the  profile.  Likewise  C  G  H  D  becomes 
a  molding,  of  which  G  H  and  C  D  are  the  miter 
lines  and  K  T  the  profile.  Therefore,  to  obtain  the 
pattern  of  the  face  pieces,  divide  the  profile  of  the 
face  K  R  into  any  convenient  number  of  spaces, 


and  from  the  points  thus  obtained  carry  lines  across 
the  face  of  the  keystone,  as  shown.  At  right  angles  in 
the  top  of  the  keystone  lay  off  a  stretchout  of  K  R,  as 
shown  by  K"  R1,  through  which  draw  the  usual 
measuring  lines.  Placing  the  T-square  parallel  to  the 
stretchout  line,  and  bringing  it  successively  against 
the  points  in  the  lines  C  D  and  A  B  bounding  the  face 
strip,  cut  the  corresponding  measuring  lines.  Then  a 
line  traced  through  these  points,  as  shown  by  C"  A"  B" 
D1,  will  be  the  pattern  for  this  part. 

In  developing  the  pattern  for  the  sink  the  usual 
method  would  be  to  divide  K  T  into  equal  spaces, 
carrying  lines  across  the  face,  and  thence  into  the 
stretchout;  but  since  this  would  result  in  confusion  of 


Pattern  Problems. 


121 


lines,  the  same  points  as  were  established  in  K  R  have 
been  used,  which  arc  quite  as  convenient  as  the  others 
mentioned,  save  that  the  points  in  K  T  must  be  ob- 
tained from  the  points  in  K  R,  bv  oiirrviiig  lines  back 
to  K  T,  as  shown,  and  in  laying  off  the  stretchout  each 
individual  space  must  be  measured  bv  the  dividers. 

K* 


K° 


S' 


Fig.  $04. — The  Patterns  for  a  Keystone  Havinij  a  Molded  Face 
with  Sink. 

At  right  angles  to  the  line  II  D  of  the  keystone  lay  off 
a  stretchout  of  K  T,  as  shown  by  K1  T',  through  the 
points  in  which  draw  the  usual  measuring  lines.  Place 
the  T-square  at  right  angles  to  the  lines  across  the  face 
of  the  keystone,  and,  bringing  it  successively  against 
the  points  in  the  lines  G  H  and  C  D,  forming  the  sides 
of  the  sink,  cut  the  corresponding  measuring  lines 
drawn  through  K'  T'.  Then  lines  traced  through  these 
points,  as  indicated  by  G1  H1  and  C"  I)1,  will  form  the 


pattern  of  the  required  sink  piece.  For  the  pattern 
of  the  piece  forming  the  sides  of  the  sink  in  the 
face  of  the  keystone,  K  R  T  becomes  the  elevation 
of  a  molding  running  in  the  direction  of  R  T,  of  which 
K  R  and  K  Tare  the  miter  lines  and  C  1)  the  profile. 
Hence,  at  any  convenient  place  above  or  below  the 
sectional  view,  lay  off  the  stretchout  of  the  line  C  D, 
as  determined  by  the  lines  drawn  across  it,  in  the  first 
operation,  all  as  indicated  by  Ca  D".  Through  the 
points  in  ('"  D"  draw  measuring  lines  in  the  usual  man- 
ner. The  next  operation,  in  course,  would  be  to  drop 
lines  from  the  points  in  the  profile  to  the  miter  lines; 
but  as  this  has  already  been  done  by  the  lines  of  the 
first  operation,  it  is  only  necessary  to  place  the  T-- 
square at  right  angles  to  the  measuring  lines,  and 
bring  it  successively  against  the  several  points  in  the 
lines  K  R  and  K  T,  and  cut  the  corresponding  measur- 
ing lines,  as  shown.  Then  a  line  traced  through  these 
points,  as  indicated  by  K3  R'  and  K3  T',  will  be  the 
pattern  of  the  piece  required. 

For  the  side  of  the  keystone,  K  L  S  R  becomes 
the  face  of  a  molding,  of  which  A  B  is  the  profile  and 
K  R  the  miter  line  at  one  side,  and  L  M  and  P  S  the 
miter  lines  at  the  other.  From  this  point  forward  the 
problem  is,  in  principle,  the  same  as  Problem  10. 
For  convenience,  and  to  avoid  confusion,  it  is  best  to 
again  make  use  of  the  same  set  of  lines  instituted  in  the 
first  part  of  the  demonstration.  Therefore,  lay  off  the 
stretchout  A'  B'  equal  to  A  B,  putting  into  it  all  the 
points  occurring  in  A  B,  through  which  draw  measur- 
ing lines  in  the  usual  manner.  Place  the  T-square  at 
right  angles  to  these  measuring  lines,  and,  bringing  it 
successively  against  the  points  in  the  line  K  R,  and 
likewise  against  L  M  and  P  S  of  the  back,  cut  corre- 
sponding measuring  lines,  as  shown.  Then  a  line 
traced  through  these  points  of  intersection,  as  shown 
by  N'  M1  L1  K4  R2  S1  P'  0',  will  be  the  outline  of  the 
required  pattern,  with  the  exception  of  that  part  lying 
between  N'  and  O',  which  make  a  duplicate  of  N  O. 
By  examination  of  the  points  in  A1  B1  and  the  lines 
drawn  through  the  same  and  making  comparison  with 
the  points  in  A  B,  it  will  be  seen  that  in  order  to  locate 
accurately  the  position  of  the  profile  of  the  window  cap 
molding  M  N  O  P,  two  additional  points,  as  shown  by  a;1 
and  '/',  have  been  introduced,  corresponding  to  x  and  y, 
the  points  of  intersection  between  the  extreme  lines  of 
the  cap  molding  itself  and  the  side  of  the  keystone 
A  B,  as  shown  in  the  elevation  by  the  curved  lines  of 
that  molding.  In  practice  it  is  frequently  necessary  to 
introduce  extra  points  in  operations  of  this  character. 


122 


The  New  Mdul    Worker  Pattern  Book. 

PROBLEM   27. 

The  Pattern  of  a  Square  Shaft  to  Fit  Against  a  Sphere. 


In  Fig.  305,  let  H  A  A1  K  be  the  elevation  of  a 
square  shaft,  one  end  of  which  is  required  to  fit  against 
the  ball  D  F  E.  Draw  the  center  line  FL,  upon  which 
locate  the  center  of  the  ball  G.  Continue  the  sides  of 
the  shaft  across  the  line  of  the  circumference  of  the 
ball  indefinitely.  From  the  points  of  intersection  be- 
tween the  sides  of  the  shaft  and  the  circumference  of 
the  ball,  A  or  A1,  draw  a  line  at  right  angles  to  the 
sides  of  the  shaft,  across  the  ball,  cutting  the  center 
line,  as  shown  at  B.  Set  the  dividers  to  G  B  as  radius, 
and  from  G  as  center,  describe  the  arc  C  C1,  cutting 
the  sides  in  the  points  C  and  C1.  Then  II  C  C1  K  will 
be  the  pattern  of  one  side  of  a  square  shaft  to  fit 
against  the  given  ball. 


Fi(j.  SOS. — The  Pattern  of  a  Square  Shaft  to  Fit  Aijainst  a  Sphere. 


PROBLEM   28. 


To  Describe  the  Pattern  of  an  Octagon  Shaft  to  Fit  Against  a  Ball. 


Let  H  F  K  in  Fig.  300  be  the  given  ball,  of  which 
G  is  the  center.  Let  D'  C"  C3  D"  E  represent  a  plan  of 
the  octagon  shaft  which  is  required  to  fit  against  the 
ball.  Draw  this  plan  in  line  with  the  center  of  the 
ball,  as  indicated  by  F  E.  From  the  angles  of  tlte 
plan  project  lines  upward,  cutting  the  circle  and  con- 
stituting the  elevation  of  the  shaft.  From  the  point  A 
or  A1,  where  the  side  in  profile  cuts  the  circle,  draw  a 
line  at  right  angles  to  the  center  line  of  the  ball  F  E, 
cutting  it  in  the  point  B,  as  shown.  Through  B,  from 
the  center  G  by  which  the  circle  of  the  ball  was 
struck,  describe  an  arc,  cutting  the  two  lines  drawn 
from  the  inner  angles  C2  C3  of  the  plan,  as  shown  at 
C  and  C1.  Then  M  C  C1  N  will  be  the  pattern  of  one 
side  of  ;.a  octagon  shaft  inhering  against  the  given 
ball  H  F  K.  If  it  be  desired  to  complete  the  eleva- 
tion of  the  shaft  meeting  the  ball,  it  may  be  done  by 
carrying  lines  from  C  and  C'  horizontally  until  they 
meet  the  outer  line  of  the  shaft  in  the  points  I);md 
D1.  Connect  C1  and  D1,  also  C  and  D,  by  a  curved 
line,  the  lowest  point  in  which  shall  touch  the  hori- 
zontal line  drawn  through  B.  Then  the  broken  line 
D  C  C1  D1  will  be  the  miter  line  in  elevation  formed 
by  the  junction  of  the  octagonal  shaft  with  the  ball. 


Fig.  S06.—The  Pattern  of  n.i  (M.iij.-,  Shaft  to  Fit  Against  n  Ball. 


123 


PROBLEM    29. 

The  Patterns  of  an  Octagonal  Shaft,  the  Profile  of  Which  is  Curved,  Fitting  over  the  Ridge  of  a  Rool. 


In  Fig.  3<l~  is  shown  tin-  elevation  :iinl  plan  of  the 
shaft  of  a  finial  of  the  design  shown  in  Kig.  :>(>S.  The 
shaft  is  octagon  throughout,  anil  if  it  were  designed  to 
stand  upon  a  level  surface,  the  method  of  obtaining  its 
patterns  would  be  the  same  in  all  respects  as  that  de- 


-M1 


Pig.  3(17.— Plan,  Elevation  and  Patterns. 


The  Patterns  of  an  Octar/imtil  Shaft,  Curved  in  Profile,  Fitting 
nri'r  tt   Rtdyp. 

scribed  in  Problem  '1\ .  As  shown  by  the  line  K /»  K, 
however,  its  lower  end  is  designed  to  tit  over  the  ridge 
of  a  roof  or  gable,  to  obtain  the  patterns  of  which  pro- 
ceed as  follows  : 

Construct  a  plan  of  the  shaft  at  its  largest  sec- 
tion, as  shown  bv  A  B  0  I)  K  K.  from  the  center  of 
which  draw  miter  lines,  as  shown  by  <1  B'and  G  F. 
Divide  the  profile  of  the  shaft  .1  L.  corresponding  to 


I'  'i  K  of  the  plan,  into  any  number  of  parts  in  the 
usual  manner,  and  from  these  points  carrv  lines  verti- 
cally crossing  the  miter  lines  (!  K  ami  <i  K.  From  the 
center  G  draw  K'  M'  at  right  angles  to  K  F,  upon  which 
line  lay  oil'  a  stretchout  of  the  profile. I  L,  drawing 

measuring  lines  through  the 
points.  Plaee  the  {"-si"1"'0 
parallel  to  the  stretchout  line, 
and,  bringing  it  successively 
against  the  points  in  G  E  and 
(1  V,  cut  corresponding  meas- 
uring lines,  as  shown,  and 
through  the  points  thus  ob- 
tained trace  lines,  all  as  indi- 
cated in  the  drawing.  This 
gives  the  general  shape  of  the 

pattern  for  the  sides  of  the  shaft.  By  inspection  of  the 
plan  and  elevation  together,  it  will  be  seen  that  to  fit  the 
shaft  over  the  roof  some  of  the  sections  composing  it 
will  require  different  cuts  at  their  lower  extremities. 
Two  of  the  sections  will  be  cut  the  same  as  the 
pattern  already  described.  They  correspond  to  the 
side  marked  A  1>  and  K  I1'  in  the  plan.  Two  others, 
indicated  in  the  plan  by  C  1)  and  II  1,  will  be  cut  to 
lit  over  the  ridge  of  the  roof,  as  shown  in  the  elevation 
by  //  ///  a.  The  remaining  four  pieces,  shown  in  plan 
bv  B  0,  D  E,  F  I  and  A  II,  will  be  cut  obliquely  to  fit 
against  the  pitch  of  the  roof,  as  shown  bv  n  o  in  the 
elevation.  For  the  sides  0  D  and  II  I,  shown  in  the 
center  of  the  elevation,  it  will  be  seen  that  the  line 
drawn  from  4  touches  the  ridge  in  the  point  m,  while 
the  line  drawn  from:!  corresponds  to  the  point  at  which 
the  side  terminates  against  the  pitch  of  the  roof. 
Therefore,  in  the  pattern  draw  a  line  from  the  center 
of  it,  on  the  measuring  line  4,  to  the  sides  of  it,  on  the 
measuring  line  :>,  all  as  shown  by  m'  n'  and  m'  n'. 
Then  these  are  the  lines  of  cut  in  the  pattern  corre- 
sponding to  in  a  and  ///  //of  the  elevation.  By  further 
inspection  of  the'  elevation,  it  will  be  seen  that  for  the 
remaining  four  sides  it  is  necessary  to  make  a  cut  in 
the  pattern  from  one  side,  in  a  point  corresponding  to 
:?  of  the  profile,  to  the  other,  in  a  point  corresponding 
to  1  of  the  profile,  all  as  shown  by  ////.  Taking  corre- 
sponding points,  therefore,  in  the  measuring  lines  of 
the  pattern,  draw  the  lines  it,'  o',  as  shown.  Then  the 
original  pattern,  modified  by  cutting  upon  these  lines, 
will  constitute  the  pattern  for  the  four  octagon  sides. 


124 


'ihe  New  Metal   Worker  Pattern  Book. 


PROBLEM   30. 

To  Construct  a  Ball  in  any  Number  of  Pieces,  of  the  Shape  of  Gores. 


Draw  a  circle  of  a  size  corresponding  to  the  re- 
quired ball,  as  shown  in  Fig.  309, which  divide,  by  any 
of  the  usual  methods  employed  in  the  construction  of 
polygons,  into  the  number  of  parts  of  which  it  is 
desired  to  construct  the  ball,  in  this  case  twelve,  all  as 
shown  by  E,  F,  Gr,  H,  etc.  From  the  center  draw 
radial  lines,  R  E  and  R  F,  etc.,  representing  the  joints 
between  the  gores,  or  otherwise  the  miter  lines.  If  the 
polygon  is  inscribed,  as  shown  in  the  illustration,  it  will 
be  observed  that  the  joint  or  miter  lines  will  lie  in  the 
surface  of  the  sphere  and  that  therefore  the  middle  of 
the  pieces,  as  shown  at  W,  C  and  «',  will  fall  inside 
the  surface  of  the  sphere  a  greater  or  less  distance 
according  to  the  number  of  gores  into  which  the. 
sphere  has  been  divided,  and  that  therefore  it  becomes 
necessary  to  construct  a  section  through  the  middle 
of  one  of  the  sides  for  use  as  a  profile  from  which  to 
obtain  a  stretchout.  It  will  be  well  to  distinguish 
here  between  absolute  accuracy  and  something  that 
will  do  practically  just  as  well  and  save  much  labor. 
This  profile,  if  made  complete,  would  have  for  its 
width  the  distance  W  u\  while  its  hight  or  distance 
through  from  R  to  a  point  opposite  would  be  equal  to 
the  diameter  of  the  circle,  or  twice  the  distance  R  U. 
As  one-quarter  of  this  section  will  answer  every  pur- 
pose, it  may  be  constructed  with  sufficient  accuracy  as 
follows  :  Supposing  R  E  F  to  be  the  piece  under  con- 
sideration, draw  a  line  parallel  to  its  center  line  R  C 
conveniently  near,  as  A  V1,  upon  which  locate  the 
points  A  and  V  by  projection  from  C  and  R,  as 
shown  by  the  dotted  lines.  From  the  point  V  erect 
the  line  B  V  perpendicular  to  V  A,  and  make  B  V 
equal  to  the  radius  of  the  circle,  or  R  V ;  then  an  arc 
of  a  circle  cutting  the  points  B  and  A  will  complete 
the  section.  This  can  be  done  by  taking  the  radius 
R  U  between  the  points  of  the  compasses  and  describ- 
ing an  arc  from  the  point  V,  whose  distance  from  V 
is  equal  to  the  distance  n'  U.  To  develop  the  pat- 
tern divide  B  A  into  any  convenient  number  of  equal 
parts,  and  from  the  divisions  thus  obtained  carry 
lines  across  the  section  P]  R  F  at  right  angles  to  a 
line  drawn  through  its  center,  and  cutting  its  miter 
lines,  all  as  shown  in  R  E  and  R  F.  Prolong  the 
center  line  R  C,  as  shown  by  S  T,  and  on  it  lay  off 
a  stretchout  obtained  from  B  A,  through  the  points 
in  which  draw  measuring  lines  in  the  usual  manner. 
Place  the  T-square  parallel  to  the  stretchout  line,  and, 


bringing  it  successively  against  ;lie  points  in  the 
miter  lines  R  E  and  R  F,  cut  the  corresponding 
measuring  lines,  as  shown.  A  line  traced  through 
these  points  will  give  the  pattern  of  a  section.  If,  on 
laying  out  the  plan  of  the  ball,  the  polygon  had  been 
drawn  about  the  circle,  instead  of  inscribed,  as  shown 
in  the  engraving,  it  is  quite  evident  that  a  quarter  of 


Fig.  309. — To  Construct  a  Ball  in  any  Number  of  Pieces,  of 
the  Shape  of  Gores. 

the  circle  would  have  answered  the  purpose  of  a  profile. 
These  points,  with  reference  to  the  profile,  are  to  be 
observed  in  determining  the  size  of  the  ball.  In  the 
illustration  presented,  the  ball  produced  will  corre- 
spond in  its  miter  lines  to  the  diameter  of  the  circle  laid 
down,  while  if  measured  on  lines  drawn  through  the 
center  of  its  sections  it  will  be  smaller  than  the  circle. 

The  patterns  for  a  ball  made  up  of  zones  or  strips 
having  parallel  sides  will  be  found  in  Section  "2  of 
this  chapter  (Regular  Tapering  Forms). 


Pattern  Problems. 

PROBLEM   31. 

The  Pattern  of  a  Round  Pipe  to  Fit  Against  a  Roof  of  One  Inclination. 


125 


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In  Fig.  310,  let  A  B  be  the  pitch  of  the  roof  and 
C  F  D  E  the  profile  of  the  pipe  which  is  to  miter 
against  it.  Let  G  0  P  H  be  the  elevation  of  the  pipe 
as  required.  Draw  the  profile  in  line  with  the  eleva- 
tion, as  shown  by  C  F  D  E,  and  divide  it  into  any  con- 
venient number  of  equal  parts.  Lay  off  a  stretchout 
in  the  usual  manner,  at  right  angles  to  and  opposite 
the  end  of  the  pipe,  as  shown  by  I  K,  and  draw  the 
measuring  Tines.  Place  the  T-square  parallel  to  the 
sides  of  the  pipe,  and,  bringing  it  successively  against 
the  divisions  of  the  profile,  cut  the  pitch  line,  as  shown 
by  A  B.  Reverse  the  T-square,  placing  it  at  right 
angles  to  the  pipe,  and,  bringing  it  successively  against 
the  points  in  A  B,  cut  the  corresponding  measuring 
lines.  A  line  traced  through  the  points  thus  obtained, 
as  shown  by  L  M  N,  will  finish  the  pattern. 


Fiij.  310— The  Pattern  of  a  Round  Pipe  to  fit.  Against  a  Roof  of 
'  One  Inclination. 

PROBLEM   32. 

The  Pattern  of  an  Elliptical  Pipe  to  Fit  Against  a  Roof  of  One  Inclination. 


In  Fig.  :.l  1,  let  N  C  D  O  be  the  elevation  of  an 
elliptical  pipe  litting  against  a  roof,  represented  by 
A  15.  Let  K  F  G  Q  be  the  section  or  profile  of  the 
pipe.  Draw  the  profile  in  convenient  proximity  to  the 
elevation,  as  shown,  and  divide  it  into  any  convenient 
number  of  equal  parts.  Place  the  T-square  parallel  to 
the  sides  of  the  pipe,  and,  bringing  it  against  the  points 
in  the  profile,  drop  lines  cutting  the  roof  line  A  B,  as 


through  the  points   in  it  draw  measuring  lines   in  the 
usual   manner.     Reverse  the  T-S(luarej  placing  it  at 


Fig.  311.— The  Pattern  of  an  Elliptical  Pipe  to  Fit  Against  a  Roof  of  One  Inclination. 

• 

shown.      Opposite  to  the  end  of  the  pipe,  and  at  right      right  angles  to  the   pipe,  and,  bringing  it  successively 
angles  to  it,  lay  off  a  stretchout,  as  shown  by  H  I,  and   ;  against  the  points   in    A    B,    cut   the   corresponding 


126 


Tin-  Ne 


1'iiti'i-n    Ik  HI/.-. 


measuring  lines,  as  indicated.       A  line    traced   through       tin-  sliurl  diameter  Q  I''  crossing  tin-  roof.    In  such  ca.-e 
these  points,  as  shown  by   l\   I.  M  .  will  lie  ilie  rei|iiirc(l       ilie  prolile  should  lie  turned    sn    that  Q  F  is  across  the 


pattern.  In  the  illustration  the  long  diameter  of  the  roof,  or  jiarallel  to  0  D,  and  the  elevation  duly  pro- 
ellipse,  or  E  G,  is  shown  as  crossing  the  roof.  The  jeete.l  from  it.  The  pipe  migbl  with  equal  facility  be 
method  of  procedure  would  be  exactly  the  same  if  the  placed 


that   the   long  diameter  should  lie  at  an 


pipe  were  placed  iu  the  opposite  position  —  that  is.  with       oblique  angle  desired. 


PROBLEM   33. 


The  Pattern  of  an  Octagon  Shaft  Fitting  Over  the  Ridge  of  a  Roof. 

In  Fig.  312,  let  A  B  C  be  the  section  and  I)  II  G.  representing  the  ridge  of  the  roof,  cut  the  corre- 
G  I  E  the  elevation  of  an  octagon  shaft  mitering  sponding  measuring  lines.  Then  a  line  traced  through 
against  a  roof,  represented  by  the  lines  F  <i  and  (!  K.  the  points  thus  obtained,  all  as  shown  by  P  ()  N"  M  L 
Place  the  section  in  line  with  the  elevation,  as  shown, 
and  from  the  angles  drop  lines,  giving  T  V  and  I 
W  of  the  elevation.  Drop  the  point  G  back  on  to 
the  section,  thus  locating  the  points  i)  and  4.  Opposite 
the  end  of  the  shaft,  and  at  right  angle's  to  it,  draw  a 
stretchout  line,  as  shown  bv  S  li,  and  through  the 
points  iu  it  draw  measuring  lines  in  the  usual  man- 
ner. Place  the  T-square  at  right,  angles  to  the  shaft, 
and,  bringing  it  successively  against  the  points  in  the 


Fig.  312.— The  Pattern  «f  an    Ortar/nn  Shaft  Fitting  Over  the  Kir/i/r  of  n  Roof. 


roof    line  formed   by   the    intersection   with    it    <>f   the      in  the  engraving,  will  be  the   lower  end  of  the  pattern 

angle  lines  in  the  elevation,  and    also  against  the  point   '•   required. 


/'illti  I'll     I'rnlil,  ///.v. 


127 


PROBLEM   34 

The  Pattern  of  a  Round  Pipe  to  Fit  Over  the  Ridge  of  a  Roof. 


Let  A   I>  (J  iii  Kin'.  :»i:;  lie  \\  section  of  tin1  roof  and 
1>  S  15  T   1']  an  elevation  of  tin.'  pipe.     1'raw  a  profile  of 
the  pipe  111    hue,   as   shown    hv    !•' 
(j    \\.      Since,     both    iiielinations 
of  the  roof  are.  to  the  same  angle, 


-\K 


I  Kit  I 

fore 


againsl    one  slope  of  the  roof,  and  lay  off  the  stretch- 
out   of   the    same    upon  the  stretchout  line  I  K,  drawn 
at  right   angles    to    the    lines   of    the  pipe,  which   may 
lie   duplicated    in  a  reverse   order   for    the  other     half, 
as  .shown.     Draw  measuring  lines  through  these  points 
in   the    usual    manner.       Place  the 
T-square    parallel    to   the  sides    of 
•   the   pipe,   and,  bringing   it  against 
the  points  in  the  prolile,  cut  the  roof 
line,  as  shown  from  B  to  T.  Reverse 
the    T-square,    placing    it    at  right 
angles  to  the  lines  of  the  pipe,  and, 
bringing  it  successively  against  the 
points  dropped  upon  the  roof  line, 
cut    the    corresponding   measuring 

/Y,/.  Jl3.-The  Pattern  ../  .1   /.'..ii.irf  Pipe  t,,  Fit  Oner  the  J}i,(,,e  ,./  .1   l!,,of.  through    the 

points,    as    shown    by   L   M    N    O 

halves  of   the   pattern  will  be  the   same.      There-   i   P,    will    form    that    end    of    the  pattern  which  meets 
space   oil'  the    half    of    the    profile    which   miters      the  roof. 

PROBLEM   35 

An  Octagon  Shaft  Mitering  Upon  the  Ridge  and  Hips  of  a  Roof. 


In  Fig.  :>  14  are  shown  the  front  and  side  elevations 
of  a  hipped  roof,  below  which  are  placed  plans,  each 
turned  so  as  to  correspond  with  the  elevation  above  it. 


Before  the  pattern  of  the  shaft  can  be  developed  it  will 
be  necessary  to  obtain  a  correct  elevation  of  its  inter- 
section with  the  roof.  Therefore,  considering  the  plan 


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FRONT  ELEVATIpN  ' 


Fir.  314.— The  Pattern  of  an  Octagon  Shaft  Fitting  Ortr  the  Ridije  and  Hips  nf  a  Rnnf. 


Aii  octagonal  shaft  is  required  to  be  mitered  down 
upon  this  roof,  so  that  its  center  line  or  axis  shall  inter- 
sect ihe  apex  of  the  roof  C,  as  shown  upon  the  plans. 


of  the  shaft  as  the  prolile  of  a  molding,  number  'all  its 
points  in  both  plans,  beginning.  For  convenience,  at  the 
ridge  of  the  roof,  and  including  the  points  where  the 


128 


The  New  Metal   Worker  Pattern  Book. 


oblique  sides  cross  the  hips  of  the  roof,  as  shown  by 
the  small  figures  1  to  11.  The  next  step  is  to  project 
lines  upward  into  the  elevations  from  each  of  these 
points,  continuing  them  till  they  intersect  the  lines  of 
the  roof,  as  shown  by  the  vertical  dotted  lines.  From 
each  of  these  intersections  in  either  view  lines  can  be 
projected  horizontally  to  the  other  view  till  they  in- 
tersect with  lines  of  corresponding  number.  Thus  the 
points  9  and  10  cut  the  line  of  the  hip  in  the  front  ele- 
vation at  the  point  B,  which,  being  carried  across  to 
the  side  view  and  intersected  with  lines  from  points  9 
and  10  from  the  plan  below  it,  give  the  correct  posi- 
tion of  those  points  in  the  side  view.  In  like  manner 
the  intersection  of  lines  from  points  tj  and  7  in  the  side 
view,  with  the  hip  line  at  D,  give  the  correct  hight  of 
those  points  in  the  front  view.  Points  5  and  8,  being 
upon  the  hips,  must  appear  in  the  elevations  at  points 
where  the  vertical  lines  from  them  cut  the  hip  lines  in 


the  elevations.  Lines  connecting  these  points  (o,  6,  7 
and  8,  and  8,  9,  10  and  11)  will  complete  the  eleva- 
tions. In  case  all  sides  of  the  roof  have  the  same  pitch 
and  the  shaft  is  a  regular  octagon,  all  the  angles  of  the 
shaft  except  ~2  and  11  will  intersect  the  roof  at  the 
same  hight,  in  which  case  it  will  only  be  necessary  to 
draw  the  front  view.  But  should  the  slope  of  the  front 
of  the  roof  be  different  from  that  of  the  sides,  it  will  be 
necessary  to  follow  the  course  above  described.  To 
develop  the  pattern,  draw  any  horizontal  line,  as  E  F, 
upon  which  place  the  stretchout  of  the  octagon  shaft 
obtained  from  the  plan,  as  shown  bv  the  small  figures, 
through  which  draw  the  usual  measuring  lines  at  right 
angles  to  it,  and  intersect  the  measuring  lines  with 
lines  of  corresponding  numbers  drawn  horizontally 
from  the  intersections  in  the  elevation.  A  line  traced 
through  these  intersections,  as  shown  by  X  Y  Z,  will 
be  the  desired  pattern. 


PROBLEM   36. 

The  Pattern  of  a  Flange  to  Fit  Around  a  Pipe  and  Over  the  Ridge  of  a  Roof. 


In  Fig.  315,  let  A  B  C  be  the  section  of  the  roof 
against  which  the  flange  is  to  fit,  and  let  O  P  S  B.  be 
the  elevation  of  the  pipe  required  to  pass  through  the 
flange.  Let  the  flange  in  size  be  required  to  extend 
from  A  to  C  over  the  ridge  B.  Since  both  sides  of  the 
roof  are  of  the  same  pitch,  both  halves  of  the  opening 
from  the  point  B  will  be  the  same.  Therefore,  for 
convenience  in  obtaining  both  halves  of  the  pattern  at 
one  operation,  the  line  B  C  may  be  continued  across 
the  pipe  toward  A1,  and  used  in  place  of  B>A,  the  dis- 
tance from  B  on  either  line  to  the  side  O  li  being  the 
same.  Under  these  conditions  it  will  be  seen  that  the 
process  of  describing  the  pattern  is  identical  with  that 
in  the  previous  problem.  Make  B  A1  equal  to  B  A, 
and  proceed  in  the  manner  described  in  the  problem 
just  referred  to.  Divide  the  profile  D  E  F  G  into  any 
number  of  equal  parts  in  the  usual  manner,  and  from 
the  points  so  obtained  carry  lines  vertically  to  the  line 
A'  C,  and  thence,  at  right  angles  to  it,  indefinitely. 
Also  carry  lines  in  a  similar  manner  from  the  points  A' 
and  C.  Draw  II  L  parallel  to  A'  C.  Make  II  I  the 
width  of  the  required  flange,  and  draw  I  K  parallel  to 
H  L.  Through  that  part  of  the  flange  in  which  the 
center  of  the  required  opening  is  desired  to  be  draw 
the  line  A*  C',  crossing  the  lines  drawn  from  the  pro- 
file. From  each  side  of  this  line,  on  the  several 


Fig.  SIS. — The  Pattern  of  a  Plunge  to  Fit  Around  a  Pipe  and  Over 
the  Ridge  of  a  Roof. 


Pattern   Problems. 


measuring  lines,  set  oil  the  same  distance  as  shown 
upon  the  corresponding  lines  between  D  F  and  I) 
E  F,  as  shown.  A  line  traced  through  the  points 
thus  obtained,  as  shown  by  D'  E1  F'  G',  will  be 


the  required  opening  to  fit  the  pipe.  Through  the 
center,  across  the  flange,  draw  the  line  N  M,  which 
represents  the  line  of  bend  corresponding  to  the  ridge 
B  of  the  section  of  the  roof. 


PROBLEM  37. 


The  Pattern  of  a  Flange  to  Fit  Around  a  Pipe  and  Against  a  Roof  of  One  Inclination. 


Let  L  M,  Fig.  316,  be  the  inclination  of 
the  roof  and  P  R  T  S.an  elevation  of  the 
pipe  passing  through  it.  N  0  then  represents 
the  length  of  the  opening  which  is  to  be  cut 
in  the  flange,  the  width  of  which  will  be  the 
same  as  the  diameter  of  the  pipe.  Let  A  B 
D  C  be  the  size  of  the  flange  desired,  as  it 
would  appear  if  viewed  in  plan.  Immediately 
in  line  with  the  pipe  draw  the  profile  G  H  I 
K,  putting  it  in  the  center  of  the  plan  of  the 
flange  A  B  D  C,  or  otherwise,  as  required. 
Divide  one-half  of  the  profile  in  the  usual 
manner,  and  carry  lines  vertically  to  the  line 
L  M,  representing  the  pitch  of  the  roof,  and 
thence,  at  right  angles  to  it,  indefinitely.  Carry 
points  in  the  same  manner  from  A  and  B. 
Draw  C'  D1  parallel  to  L  M.  Make  C1  A1  equal 
to  A  C,  or  the  width  of  the  required  flange, 
and  draw  A1  B1  parallel  to  C1  D1.  Then  C1  A1 
B1  D1  will  be  the  pattern  of  the  "required  flange. 
Draw  E1  F1  through  it  at  a  point  corresponding 
to  E  F  of  the  plan,  crossing  the  lines  drawn 
from  the  profile.  From  E1  F'  set  off  on  each 
side,  on  each  of  the  measuring  lines  crossing  it, 
the  width  of  opening,  as  measured  on  corre- 
sponding lines  of  the  plan,  measuring  from  E 
F  in  the  plan  to  the  profile.  Through  the 
points  thus  obtained  draw  a  line,  which  will 
give  the  shape  of  the  opening  to  be  cut,  all  as 
shown  by  G1  H1  I1  K1. 


Fig.  316  —The  Pattern  of  a  Flange  to  Fit  Around  a  Pipe  and 
Against  a  Roof  of  One  Inclination. 


130 


Tlie  New  Metal    Worker  Pattern   Book. 

PROBLEM   38. 


The  Pattern  for  a  Two-Piece  Elbow. 


In  Fig.  317,  let  A  C  B  D  be  the  profile  of  the  pipe 
in  which  the  elbow  is  to  be  made.  Draw  an  elevation 
of  the  elbow  with  the  two  arms  at  right  angles  to  each 
other,  one  of  which  is  projected  directly  from  the  pro- 
file, as  shown  by  E  G  I  H  K  F.  Draw  the  diagonal 
line  G  K,  which  represents  the  joint  to  be  made. 
Divide  the  profile  into  any  convenient  number  of  equal 
parts.  Place  the  T-square  parallel  to  the  lines  of  the 
arm  of  the  elbow,  opposite  the  end  of  which  the  profile 
has  been  drawn,  and,  bringing  the  blade  successively 
against  the  several  pojnts  in  the  profile,  drop  corre- 
sponding points  on  the  miter  or  joint  line  K  G,  as  shown 
by  the  dotted  lines.  Opposite  the  end  of  the  same 
arm,  and  at  right  angles  to  it,  lay  off  a  stretchout  line, 
M  N,  divided  in  the  usual  manner,  and  through  the 
divisions  draw  measuring  lines,  as  shown.  Place  the 
blade  of  the  T-square  at  right  angles  to  the  same  arm 
of  the  elbow,  or,  what  is  the  same,  parallel  to  the 
stretchout  line,  and,  bringing  it  successively  against 
the  points  in  K  G,  cut  the  corresponding  measuring 
lines,  as  shown.  A  line  traced  through  these  points, 
as  indicated  by  K  P  O,  together  with  M  N,  will  form 
the  required  pattern. 


Fig.  317.— The  Pattern  for  a  Two-Piece  Elbow, 


PROBLEM  39. 


The  Patterns  for  a  Two-Piece  Elbow  in  an  Elliptical  Pipe.— Two  Cases. 


The  only  difference  to  be  observed  in  cutting  the 
patterns  for  elbows  in  elliptical  pipes,  as  compared  with 
the  same  operations  in  connection  with  round  pipes, 
lies  with  the  profile  or  section.  The  section  is  to  be 
placed  in  the  same  position  as  shown  in  the  rules  for 
cutting  elbows  in  round  pipe,  but  it  is  to  be  turned 
broad  or  narrow  side  to  the  view,  as  the  requirements 
of  the  case  may  be.  In  Figs.  318  and  319  are  shown 
elevations  and  profiles  of  two  right  angled  two-piece 
elbows  in  elliptical  pipes.  In  Fig.  318  the  broad  side 
of  the  ellipse  is  presented  to  view,  while  Fig.  319 
shows  the  narrow  side,  as  indicated  by  the  respective 


positions  of  the  profiles.  Although  the  results  in  the 
two  cases  are  different  in  consequence  of  the  position 
of  the  profiles,  the  method  of  procedure  is  exactly  the 
same.  Similar  parts  in  the  two  drawings  have  been 
given  the  same  reference  letters  and  figures,  so  that  the 
following  demonstrations  will  apply  equally  well  to 
either :  Let  A  C  E  F  D  B  be  the  elevation  of  the 
elbow  and  II  G  K  I  its  section.  Draw  C  D,  the 
miter  line.  Divide  the  profile  i:;  the  usual  manner, 
as  indicated  by  the  small  figures,  and  by  means 
of  the  T-square  placed  parallel  lo  the  "arm,  drop 
points  upon  the  miter  line,  as  shown.  Opposite  the, 


Pattern    Problems. 


131 


end  of  the  arm  lay  oil'  a  stretchout,  M  N,  and  through 
tin-    points    in    it    draw    the    usual    measuring    lines. 


points  in  the  miter  line,  cut  the  corresponding  meas- 
uring    lines.      A    line    traced    through    these    points, 


Fig.  318. 


Fig.  319. 


A  Two-Piece  Elbow  in  Elliptical  Pipe. 


liVverse   the   T-square,    placing  it  at  right    angles  to 
the  arm,  and,  bringing  it  in  contact   with  the  several 


as   shown    by   L  P  O,    will    constitute   the   required 


miter. 


PROBLEM  40. 


The  Patterns  for  a  Three-Piece  Elbow. 


In  Fig.  320,  let  E  M  L  I  H  K  N  F  be  the  ele- 
vation of  a  three-piece  elbow.  The  drawing  of  a 
three-piece  elbow,  at  any  angle  whatever,  should  be 
so  constructed  that  the  middle  section  or  portion  bears 
the  same  angle  with  reference  to  the  two  arms. 
Since  the  two  arms  in  the  present  instance  are  at  right 
angles  (90  degrees)  to  each  other,  the  middle  section 
must  therefore  be  drawn  at  an  angle  of  45  degrees  to 


both.  Make  its  diameter  the  same  as  that  of  the  two 
.arms,  and  draw  the  miter  lines  M  N  and  L  K.  Draw 
the  profile  A  B  C  in  line  with  the  arm  from  which  the 
pattern  is  to  be  taken,  as  shown,  and  divide  it  into  any 
convenient  number  of  equal  parts.  Place  the  blade 
of  the  T-square  parallel  to  this  arm  of  the  elbow,  and, 
bringing  it  against  the  points  in  the  profile,  drop  cor- 
responding points  upon  the  miter  line  L  K.  At  right 


132 


The  Xew   Metal    Worker    Pattern    Book. 


angles  to  L  I  draw  a  stretchout,  as  R  S,  through  the 
divisions  in  which  draw  measuring  lines  in  the  usual 
manner.  Placing  the  T-square  at  right  angles  to  L  I, 
and  bringing  it  successively  against  the  points  in  the 
miter  line  L  K,  cut  the  corresponding  measuring  lines, 


drop  like  divisions  upon  M  X.  At  right  angles  1<>  1. 
M  lay  off  a  stretchout  of  the  profile  A  15  0,  as  shown 
by  P  0,  through  the  points  in  which  draw  measuring 
lines  in  the  usual  manner.  Reversing  the  position  of 
the  set-square  so  that  its  long  side  shall  come  at  right 


Fig.  310.— A  Three-Plece  Elbow. 


as  shown.  Then  the  line  T  U  V,  traced  through  the 
points  thus  obtained,  forms  in  connection  with  S  R 
the  pattern  of  an  end  section. 

Place  the  45-degree  set  square  against  the  blade 
of  the  T-square  so  that  its  oblique  or  long  side  shall 
coincide  with  the  lines  of  the  middle  section  of  the 
elbow,  and,  bringing  it  against  the  points  in  L  K, 


angles  to  M  L,  or,  what  is  the  same,  parallel  to  the 
stretchout  line,  bringing  it  successively  against  the 
several  points  in  the  miter  lines  M  N  and  L  K,  and 
cut  the  corresponding  measuring  lines.  Then  lines 
traced  through  these  points,  as  shown  by  D  X 
Y  and  G  W  Z,  will  be  the  pattern  of  the  middle 
section. 


PROBLEM  41. 


The  Patterns  for  a  Four-Piece  Elbow. 


In  constructing  the  elevation  of  a  four-piece 
elbow,  first  draw  the  profile  A  B  C,  from  which  pro- 
ject one  of  the  arras  of  the  elbow,  as  shown  by  the  lines 
A  F  and  C  G,  Fig.  321,  At  right  angles  to  this  lay 


off  the  other  arm  of  the  elbow.  M  T.  X  T.  continuing 
the  lines  of  each  until  they  intersect.  Through  the 
points  of  intersection  draw  the  diagonal  line  a  d. 
Establish  the  point  a  on  this 


diagonal  line   at  con- 


Pattern     Problems. 


133 


venienee,  and  from  it.  draw  the  lines  <t  I  and  a  c.  at 
right,  angles  to  the  two  arms  of  the  elbow  respectively. 
From  ii  as  center,  and  with  a  I*  as  radius,  describe  the 
are  l>  f  e  r,  as  shown,  which  divide  into  three  equal 
parts,  thus  obtaining  the  points- /and  e.  Through  /' 


w 


Fig.  SSI.— A  Four-Piece  Elbow. 

and  e,  to  the  center  a,  draw  the  lines  f  a  and  e  «, 
which  will  represent  the  centers  of  the  middle  sec- 
tions of  the  elbow,  at  right  angles  to  which  the  sides 
of  the  same  are  to  be  drawn.  Through/,  and  at  right 


angles  to /a,  draw  L  K,  meeting  M  L  in  the  point  L, 
and  stopping  on  the  line  </  '/  at  the  point  K.  Through 
'•,  and  at  right  angles  to  e  a,  draw  a  line,  commenc- 
ing in  the  point  K  and  terminating  in  G  where  it 
meets  the  line  E  (1.  In  like  manner  draw  the  lines 
of  the  inner  side  of  the  elbow,  as  shown  by  F  H 
and  H  I.  Draw  the  miter  or  joint  lines  F  G,  H  K 
and  L  1,  as  shown.  For  the  patterns  proceed  as  fol- 
lows :  Divide  the  profile  into  any  convenient  number 
of  equal  parts.  Place  the  T-square  parallel  to  E  G, 
and,  bringing  the  blade  against  the  points  in  the  pro- 
file, drop  corresponding  points  upon  the  miter  line  F 
G.  Change  the  T-square  so  that  its  blade  shall  be 
parallel  to  the  lines  of  the  second  section  of  the  elbow, 
and,  bringing  it  against  the  points  in  F  G,  cut  corre- 
sponding points  on  II  K.  Opposite  the  end  of  and 
at  right  angles  to  the  lower  arm  of  the  elbow,  lay  off 
the  stretchout  line  O  P,  as  shown,  through  the  divi- 
sions in  which  draw  the  usual  measuring  lines.  Place 
the  T-square  at  right  angles  to  the  arm  of  the  elbow, 
and,  bringing  it  successively  against  the  points  in  the 
miter  line  F  G,  cut  the  corresponding  measuring  lines. 
Then  a  line  traced  through  the  points  thus  obtained, 
as  shown  from  R  to  T,  will  with  O  P  constitute 
the  pattern  of  one  of  the  arms.  Produce  a  e,  repre- 
senting the  middle  of  the  second  section  in  the  elbow, 
as  shown  by  V  "W,  upon  which  lay  off  a  stretchout, 
and  through  the  points  in  the  same  draw  measuring 
lines.  Placing  the  T-square  parallel  to  a  e,  or,  what  is 
the  same,  at  right  angles  to  the  second  section  of  the 
elbow,  bring  it  against  the  several  points  in  the  miter 
lines  II  K  and  F  G,  and  cut  the  corresponding  measur- 
ing lines.  Then  lines  traced  through  the  points  thus 
obtained,  as  shown  from  X  to  Z  and  Y  to  S,  will  give 
the  pattern. 


PROBLEM    42. 
The  Patterns  for  a  Five-Piece  Elbow. 


To  construct  the  elevation  of  a  five-piece  elbow, 
first  draw  the  profile,  as  ABC,  Fig.  322,  from  which 
project  one  of  the  arms  of  the  elbow,  as  shown  at  the 
left  bv  K  S  H  0,  continuing  its  lines  indefinitely.  At 
right  angles  to  this  lay  off  the  other  arm,  continuing  its 
lines  till  they  intersect  those  of  the  horizontal  arm,  or 
till  their  outer  lines  intersect,  as  at  g.  Draw  g  a  at  an 
angle  of  45  degrees  to  either  arm,  upon  which  establish 
the  point  a  with  reference  to  the  curve  which  it  is  de- 


sired the  elbow  shall  have,  and  from  it,  at  right  angles 
to  the  two  arms  of  the  elbow  respectively,  draw  a  b  and 
a  c.  From  a  as  center,  with  ab  as  radius,  describe  the 
nrcbfedc,  which  divide  into  four  equal  parts,  thus 
obtaining  the  points  d,  e  and/  and  draw  d  a,  c  a  and 
/  a.  Then  these  lines  represent  center  lines  of  the  sev- 
eral sections  of  which  the  elbow  is  composed,  at  right 
angles  to  which  their  sides  are  to  be  drawn. 

It   may   be    here   remarked   that    the    number    of 


134: 


The   \/-ii>   Mi'tnl     \Yoi-L-'' i-    I'altfrii    JSouk. 


center  lines  made  use  of  in  dividing  the  quarter  circle 
I  c  represents  the  number  of  pieces  in  the  elbow. 
Therefore,  to  draw  an  elevation  of  an  elbow  in  any 
number  of  pieces,  construct  the  quadrant  ale  as  above 
described,  then  divide  b  c  into  such  a  number  of  parts 
that  the  number  of  lines  drawn  to  a  (including  a  b  and 
a  c)  shall  equal  the  number  of  pieces  required.  Thus 
the  five  lines  a  6,  a/,  a  c,  a  d  and  a  c  are  the  center 
lines  of  the  five  pieces  of  which  the  elbow  shown  in 
Fig.  322  is  constructed.  Although  the  two  extreme 
lines  a  b  and  a  c  are  not,  strictly  speaking,  center  lines, 
their  relation  to  the  adjacent  miter  lines  is  the  same  as 
that  of  the  other  lines  radiating  from  a.  Through/, 
and  at  right  angles  to  fa,  draw  V  S,  joining  the  side 
of  the  arm  E  S  in  the  point  S,  and  joining  a  corre- 
sponding line  drawn  through  e  in  the  point  V.  In  like 
manner  draw  the  line  T  E,  representing  the  inner  side 
of  the  same  section.  The  remaining  sections  are  to  be 
obtained  in  the  same  way.  As  but  one  section  is 
necessary  for  use  in  cutting  the  patterns,  the  others 
may  or  may  not  be  drawn,  all  at  the  option  of  the 
pattern  cutter.  Draw  the  miter  or  joint  lines  S  E, 
V  T,  etc.  Divide  the  profile  (or  one-half  of  it)  in  the 
usual  manner.  Place  the  T-square  parallel  to  the  lines 
of  the  arm,  and,  bringing  the  blade  against  the  several 
points  in  the  profile,  drop  corresponding  points  upon 
the  miter  line  S  E.  Shift  the  T-square  so  that  the 
blade  shall  be  parallel  to  the  part  V  S  E  T,  and  trans- 
fer the  points  in  S  E  to  V  T,  as  shown.  For  the  pat- 
tern of  the  arm,  at  right  angles  to  it  lay  off  a  stretch- 
out of  A  B  C,  as  shown  by  F  G,  through  the  points  in 
which  draw  the  usual  measuring  lines.  Place  the 
T-square  at  right  angles  to  the  arm,  and,  bringing  it 
against  the  points  in  E  S,  cut  the  corresponding  meas- 
uring lines,  as  shown.  Then  a  line  traced  through 
these  points,  as  shown  from  H  to  I,  will  be  the  pattern. 
For  the  pattern  of  the  piece  S  V  T  E  prolong  the  line 
a/,  as  shown  by  L  K,  upon  which  lay  off  a  stretchout, 


through  the  points  in  which  draw  the  measuring  lines 
in  the  usual  manner.  Placing  the  T-square  at  right, 
angles  to  S  T,  or,  what  is  the  same,  parallel  to  the 
stretchout  line,  bring  it  against  the  several  points  in 


Fig.  SSS  —A  Five-Piece  Elbow. 

the  lines  E  S  and  T  V,  and  cut  the  corresponding 
measuring  lines.  Then  lines  traced  tli rough  the  points 
thus  obtained,  all  as  shown  by  N  P  0  M,  will  be  the 
pattern  sought. 


PROBLEM  43. 

The  Patterns  for  a  Pipe  Carried  Around  a  Semicircle  by  Means  of  Cross  Joints. 


In  Fig.  323,  let  F  E  D  be  the  semicircle  around 
which  a  pipe,  of  which  A  C  B  is  a  section,  is  to  be 
carried  by  means  of  any  suitable  number  of  cross 
joints,  in  this  instance  ten.  Divide  the  semicircle  F  E 
D  into  the  same  number  of  equal  parts  as  there  are  to 
be  joints,  which,  as  just  stated,  is  to  be  ten,  all  as 


shown  by  D,  0,  P,  E,  S,  E,  etc.,  and  draw  lines  from 
each  point  to  Z.  As  there  are  to  be  ten  joints  there 
must  necessarily  be  eleven  pieces,  therefore,  according 
to  the  directions  given  in  the  previous  problem,  the 
semicircle  must  be  divided  into  such  a  number  of 
equal  parts  that  the  number  of  lines  radiating  from  Z 


Pntlvnt 


135 


shall  be  eleven,  all  as  shown,  each  line  serving  as  the 
center  line  <>f  a  piece.  From  L>  toward  the  center  Z 
set  off  the  diameter  of  the  pipe  A  B.  as  shown  l>v  tlie 
point  A'.  From  Z  as  eenter,  with  the  radius  Z  A1, 
draw  the  dotted  line  representing  the  inner  line  of  the 
pipe,  and  cutting  the  radial  lines  previously  drawn  in 
the  points  0',  P1,  etc.  Through  0  and  0'  draw  lines 
at  right  angles  to  0  Z  and  continue  them  in  either 
direction  till  they  intersect  with  the  lines  drawn 
through  P  and  P1  on  the  one  side  and  through  D  and 
A'  on  the  other.  Each  pair  of  lines  is  to  be  drawn  at 


corresponding  to  the  full  sections  composing  the  body 
of  the  pipe.  For  the  pattern  of  the  end  section  pro- 
ceed as  follows :  Divide  the  profile  A  C  B  in  the  usual 
manner  into  anv  convenient  number  of  equal  parts, 
and  from  the  points  thus  obtained  carry  lines  upward 
at  right  angles  to  Z  D.  cutting  T1  T.  Prolong  the 
line  Z  D.  and  upon  it  place  a  stretchout  from  the 
profile  A  C  B,  perpendicular  to  which  draw  measur- 
ing lines  in  the  usual  manner.  With  the  T-square 
placed  parallel  to  Z  D,  and  brought  successively 
against  the  points  in  T'  T,  cut  the  measuring  lines  of 


Fig.  StS.—A  Pipe  Carried  Around  a  Semicircle  by  Means  of  Cross  Joints. 


right  angles  to  its  respective  radial  or  center  line. 
Through  the  points  of  intersection  draw  the  lines  T  T1, 
U  U',  etc.,  which  will  represent  the  lines  of  the  joints 
or  miters. 

It  will  appear  by  inspection  that  the  point  U  is 
equidistant  from  P  and  O,  and  that  U'  is  also  equidis- 
tant from  P'  and  0',  and  that  therefore  the  lines  U  U', 
T  T1,  etc.,  if  continued  inward  must  arrive  at  the 
center  Z.  Thus  the  joint  lines,  like  the  center  lines, 
must  radiate  from  the  center  of  the  semicircle. 

Draw  the  profile  of  the  pipe  A  C  B  directly  below 
and  in  line  with  one  end  of  the  pipe,  all  as  shown  in 
the  engraving.  As  may  be  seen  by  inspection  of  the 
diagram,  two  patterns  are  required,  one  corresponding 
to  the  half  section  occurring  at  the  end,  and  the  other 


corresponding  numbers.  Then  a  line  traced  through 
the  points  of  intersection  thus  obtained,  as  shown  by 
I  K  L,  will  be  the  shape  of  the  miter  cut,  and  G  I  K 
L  H  will  be  the  complete  pattern  for  one  of  the  end 
pieces.  For  the  pattern  of  one  of  the  large  pieces, 
as  U  V  V  U',  lay  off  a  stretchout  of  A  C  B  upon  its 
center  line  extended,  as  shown  by  M  N,  and  through, 
the  points  in  it  draw  measuring  lines  in  the  usual  man- 
ner. Place  the  T-square  parallel  to  U  V  and,  bringing 
it  against  the  points  in  U  U1,  cut  the  line  V  V.  Next 
place  the  T'S(luare  parallel  to  the  stretchout  line, 
and,  bringing  it  against  the  several  points  in  the  miter 
lines  U1  U  and  V  V,  cut  the  corresponding  meas- 
uring lines,  all  as  shown,  thus  completing  the  pat- 
tern. 


136 


The    New    Metal    Worker    Pattern    Book. 


PROBLEM   44. 
The  Patterns  for  an  Elbow  at  Any  Angle. 


LetDFHKLIGE  in  Fig.  324  be  the  eleva- 
tion, of  a  pipe  in  which  elbows  are  required  at  special 
angles.  In  convenient  proximity  to  and  in  line  with 


Fig.  Sij.—An  Elbow  at  Any  Angle. 

one  end  of  the  pipe  draw  a  profile,  as  shown  by  A  B 
C,  which  divide  in  the  usual  manner.  Placing  the 
T-square  parallel  to  the  first  section  of  the  pipe,  and, 
bringing  it  against  the  several  points  in  the  profile, 
drop  corresponding  points  upon  F  G.  Shift  the 


J-square,  placing  it  parallel  to  the  second  section,  and, 
bringing  it  against  the  several  points  in  F  G,  drop 
them  upon  H  I.  At  right  angles  to  the  first  section 

lay  off  a  stretchout  of  A  B  C, 
as  shown  by  T  U,  through  the 
points    in     which     draw     the 
customary    measuring     lines. 
Placing  the  T-square  at  right 
angles   to  this  section  of  the 
pipe,  and  bringing  it  against 
the  several  points  in  F  G,  cut 
the   corresponding  measuring 
lines.     Then  the  line  ELS 
traced    through    these    points 
will,  with  the  line  T  U,  be  the 
pattern  sought.     The  pattern 
for  the  opposite  end  is  to  be 
obtained   in   like  manner,   all 
as  shown  by  M  N  0  P,   and 
therefore  need  not  be 
described  in    detail. 
For    the    pattern    of 
the    middle    section 
lay  off  a  stretchout, 
W  V,  at  right  angles 
to  it,  with  the  cus- 
tomary, measuring 
lines.      Placing   the 
T-square    at     r  i  gh  t. 

angles  to  the  section,  bring  it  successively  against 
the  points  in  G  F  and  I  H,  and  cut  the  correspond- 
ing measuring  lines,  as  shown.  Then  lines  traced 
through  these  points,  as  shown  by  Y  X  Z  Q,  will 
be  the  pattern  sought.  The  positions  of  the  longi- 
tudinal joints  in  the  several  sections  of  this  elbow,  as 
well  as  those  of  all  others,  are  determined  by  the 
order  in  which  the  measuring  lines  drawn  through 
the  stretchout  are  numbered.  In  the  present  instance 
the  joints  are  allowed  to  come  on  the  back  of  the 
pipe,  or,  in  other  words,  upon  D  F  II  K,  which 
corresponds  to  the  point  1  in  the  profile.  Hence, 
in  numbering  the  measuring  lines  in  the  several  stretch- 
outs, point  1  is  placed  at  the  commencement  and 
ending,  while  if  it  were  desired  to  have  the  joint 
in  either  piece  come  on  the  opposite  side,  or  at  a  point 
corresponding  to  9  of  the  profile,  the  stretchout  would 
have  commenced  and  ended  with  that  figure,  the 


Pattern    Problems. 


137 


figure  1  in  that  case  coming,  in  regular  order,  where  9 
now  occurs.  The  effect  of  such  u  change  upon  anv  of 
the  patterns  here  given  would  be  the  same  as  if  they 


were  cut  in  two  upon  the  line  9  and  the  two  halves 
were  transposed. 


PROBLEM   45. 

The  Patterns  for  a  Bifurcated  Pipe,  the  Two  Arms  Being  the  Same  Diameter  as  the  Main  Pipe,  and  Leaving:  It  at 

the  Same  Angle. 


In  Fig.  325  is  shown  an  elevation  of  a  bifurcated 
pipe,  all  arms  being  of  the  same  diameter.  In  this 
problem,  as  in  many  others,  it  becomes  necessary  to 
first  make  a  correct  drawing  of  the  intersection  of  the 
parts  showing  the  miter  lines  correctly ;  after  which 


of  the  miter  lines  will  at  once  be  determined.  Thus 
the  intersection  of  the  three  bisecting  lines  at  E  gives 
the  point  at  which  the  miter  lines  starting  from  the 
points  P,  E  and  K  must  meet. 

In  line  with  the  upper  end  of  the  pipe  draw  a 


Fig.  MS.— The  Pattern  for  a  Bifurcated  Pipe. 


the  method  of  laying  out  of  the  miter  patterns  is  the 
same  as  that  employed  in  several  other  problems 
immediately  preceding  this.  If.  in  this  case,  each  arm 
of  the  pipe  be  divided  longitudinally  into  two  equal 
parts,  as  shown  by  the  center  lines,  and  each  half  be 
considered  as  a  separate  molding  the  correct  position 


profile  of  it,  as  shown  by  A  C  B.  A  profile  will  also 
be  needed  in  one  of  the  oblique  arms,  a  half  only 
being  shown  at  A'  C'  B'  on  account  of  the  limited 
space.  For  the  pattern  of  the  upper  portion  of  the 
pipe,  divide  the  profile  A  C  B  into  any  number  of 
equal  spaces,  and  place  the  stretchout  of  the  same  on 


138 


The    New    Metal    Worker   Pattern    Book. 


any  line  drawn  at  right  angles  to  S  P,  as  shown  by 
the  continuation  of  S  D  to  the  left,  and  draw  the 
usual  measuring  lines.  Next  drop  the  points  from 
the  profile  A  C  B  parallel  with  S  P  till  they  cut 
the  miter  line  PEE;  then  placing  the  T-square  at 
right  angles  to  S  P,  drop  the  points  from  the  miter 
line  P  E  into  measuring  lines  of  corresponding  num- 
ber. A  line  traced  through  these  points  of  intersection, 
as  shown  from  E"  to  P',  will  give  the  miter  cut  on  the 
lower  end  of  the  pipe  S  D  E  P,  one-half  of  which  only 
is  shown  in  the  engraving.  The  pattern  for  the  piece 
E  F  J  K  is  obtained  in  exactly  the  same  manner,  and 
might  be  obtained,  so  far  as  the  half  indicated 


by  C'  B'  on  profile  is  concerned,  from  the  original 
profile,  by  simply  continuing  the  lines  through  to  the 
miter  line  J  F,  as  shown.  For  simplicity,  therefore, 
the  profile  A'  C'  B'  is  divided  into  the  same  number 
of  equal  parts  as  the  original  profile,  and  a  stretchout  of 
it  is  placed  upon  any  line,  as  T  U.  drawn  at  right  angles 
to  E  F.  The  points  are  then  dropped  from  the  profile 
both  ways,  cutting  the  miter  lines  K  R  E  and  J  F, 
after  which,  with  the  T-square  placed  parallel  to  T  U, 
they  can  be  dropped  into  the  measuring  lines  of  the 
stretchout.  Lines  traced  through  the  points  of  inter- 
section will  constitute  the  required  pattern,  as  shown 
by  K'  R'  E'  R"  K"  X  W  V. 


PROBLEM  46. 


The  Patterns  for  the  Top  and  Bottom  of  a  "  Common "  Skylight  Bar. 


In  Fig.  326,  A  B  represents  a  portion  of  the  pro- 
file of  the  ridge  bar,  or  of  the  ventilator  forming  the 
top  finish  of  a  skylight,  against  which  the  upper  end 
of  a  "common"  bar  is  required  to  miter;  and  C  D 
represents  the  profile  of  the  curb  or  finish  against 
which  the  lower  end  of  the  bar  miters.  The  parallel 
oblique  lines  connecting  the  two  show  the  side  eleva- 
tion of  the  bar  whose  profile  is  shown  at  E  F  F1. 

As  the  profile  consists  of  two  symmetrical  halves, 
either  half,  as  E  F  or  E  F1,  may  be  chosen  to  work 
from,  and  as  it  contains  no  curved  portions  it  is 
simply  necessary  to  number  all  of  its  points  or  angles, 
and  then  to  place  a  complete  stretchout  of  the  same 
upon  any  line  drawn  at  right  angles  to  the  lines  of  the 
molding,  as  Gr  H,  and  to  draw  the  usual  measuring 
lines,  all  as  shown.  As  a  properly  drawn  elevation 
shows  the  intersection  of  the  points  of  the  profile  with 
the  two  miter  lines  A  B  and  C  D1,  it  is  only  neces- 
sary to  place  the  T-square  parallel  to  the  stretchout 
lines  G  H,  and  bring  it  successively  against  the  points 
in  A  B  and  C  D1,  and  cut  corresponding  measuring 
lines,  as  shown  at  I  J  and  K  L.  Straight  lines  con- 
necting the  points  of  intersection  will  complete  the 
pattern,  as  shown  at  I  J  L  K.  The  length  of  the 
pattern,  which  is  here  shown  indefinite,  must  be 
determined  by  a  detail  drawing,  in  which  the  rise  M 
B  and  the  run  M  D1  are  correctly  given. 

The  patterns  for  the  "jack"  bar    and    for    the 


"hip  "  bar  will  be  given  later  among   those  problems 
in  which  the  development  of  the  miter  line  and  the 


Fig.  32fi.—The  Patterns  for  a  "  Common  "  Skylight  Bar. 

raking  of   the  profile  are  necessary,  with  which  they 
are  properly  classed. 


1'atkrn     I'rMems. 

PROBLEM   47. 


139 


The  Patterns  for  a  T-Joint  Between  Pipes  of  the  Same  Diameters. 


Let  D  F  G  II  M  F  K  ft  in  Fig.  327  be  the  eleva- 
tion of   two   ]>i])cs  of  the    same   size  meeting  at  right 


Fig.  327.— A  T-  Joint  between  Pipes  of  the  Same  Diameter. 

angles  and  forming  a  T,  of  which  A  B  C  D  and  A'  B' 
(J1  D'  are  profiles  drawn  in  line  with  either  piece.  As 
the  two  profiles  are  alike,  and  as  the  end  of  one  piece 
(D  E  F  K)  comes  against  the  side  of  the  other  piece 
(G  I  M  H),  both  halves  of  D  E  F  K,  B  A  D  and  B  C 


D,  will  miter  with  one-half,  B'  C1  D',  of  the  piece  G  I 
M  H.  By  projecting  the  points  B  and  B'  from  the 
profiles  through  their  respective  elevations  the  point  L 
is  found,  which  being  connected  with  the  points  F  and 
K  gives  the  miter  lines.  Space  the  profile  A  B  C  D 
into  any  number  of  equal  parts  and  lay  off  the  stretch- 
out 1ST  O  ut  right  angles  to  the  pipe  of  which  ABC 

D  is  the  profile,  as  shown, 
through  the  points  in 
which  draw  the  usual 
measuring  lines.  Set  the 
T-square  at  right  angles 
to  this  pipe,  and,  bring- 
ing the  blade  against 
the  several  points  on 
the  rniter  lines,  cut  the 
corresponding  measuring 
lines  drawn  through  the 
stretchout,  as  indicated 
by  the  dotted  lines.  Then 
N  F1  U  V  W  O  will  be 

the  pattern  for  the  upper  piece.  As  both  halves  of 
this  piece  (dividing  now  upon  the  line  A  C)  will  be 
alike  only  one-half  of  the  profile  (A  B  C)  has  been 
divided,  but  the  stretchout  is  made  complete.  For 
the  pattern  of  the  other  piece,  divide  its  profile  into 
any  convenient  number  of  equal  parts  and  lay  off 
the  stretchout  on  the  line  R  T,  drawn  at  right  angles 
to  the  pipe.  Placing  the  T-square  parallel  with 
the  pipe  drop  points  upon  the  miter  lines  from  that 
portion  of  the  profile  (B1  C'  D')  which  comes  in  line 
with  them ;  then  place  the  blade  of  the  T-square  at 
right  angles  to  the  pipe,  and,  bringing  it  against  the 
several  points  in  the  miter  lines,  cut  the  corresponding 
measuring  lines,  as  shown  by  the  dotted  lines.  A 
line,  X  Y  Q  Z,  traced  through  these  points  will 
bound  the  opening  to  be  cut  in  the  pattern  for  the 
lower  pipe.  From  the  points  1  in  the  stretchout 
draw  the  lines  R  P  and  T  S,  in  length  equal  to  the. 
length  of  the  pipe.  Connect  P  S.  Then  P  R  T  S 
will  be  the  required  pattern.  The  seam  iji  the  pipe 
may  be  located  as  shown  in  the  engraving,  or  at  some 
other  point,  at  pleasure. 


140 


The   New    Metal    Worker    Pattern    Book. 

PROBLEM   48. 


The  Patterns  for  a  Square  Pipe  Describing  a  Twist  or  Compound  Curve. 


As  problems  of  this  nature  frequently  occur  in 
connection  with  hot  air  pipes,  grain  chutes,  etc., 
this  problem  is  given  as  embodying  principles  which 
can  often  be  made  use  of.  The  upper  opening  of  the 
pipe  in  this  case  is  required  to  be  in  a  horizontal 
plane,  while  the  lower  opening  is  in  a  vertical  position 
and  placed  at  a  given  distance  below  and  to  one  side 
of  the  top,  the  pipe  describing  a  quarter  turn  when 
viewed  from  either  the  top  or  the  front. 

To  more  fully  illustrate  the  nature  of  the  prob- 
lem, a  perspective  view  of  it  is  shown  in  Fig.  328,  in 
which  the  pipe  is  represented  as  being  contained 
within  a  cubical  shaped  solid.  The  solid,  of  which 
the  pipe  is  represented  as  forming  a  part,  is  shown  in 
outline,  the  pipe  itself  being  shaded  to  show  its  form, 
while  upon  the  front  and  lower  side  of  the  solid  are 
shown  in  dotted  lines  the  front  elevation  and  plan  of 
the  pipe.  Thus  G  F  T  C  represents  the  front  view  of 
a  solid  just  large  enough  to  contain  the  pipe,  in  which 
A  B  C  D  shows  the  position  of  the  lower  opening,  and 
A  B  E  FD  C  shows  the  curve  of  the  pipe  as  seen  from 
the  front.  G  II  S  B1  is  the  top  of  tin-  solid  in  which 
the  upper  opening  N  P  S  R  is  situated.  The  curve  of 
the  pipe  in  plan  has  been  projected  to  the  lower  face 
of  the  solid  by  vertical  lines,  R  L,  and  others  not 
shown,  and  is  shown  by  C  J  K  L  M  D.  To  state  the 
case  simply,  then,  ABE  is  the  profile  of  the  piece  of 
metal  forming  the  top  of  the  pipe,  while  1)  M  L  and 
C  J  K  are  the  two  miter  lines,  or  the  plans  of  the 
intersecting  surfaces,  and  CDF  is  the  profile  of  the 
lower  side  of  the  pipe  intersecting  the  same  miter 
lines.  The  top  and  bottom  pieces  being  developed, 
it  is  only  necessary  to  reverse  the  operation  and  con- 
sider the  lines  of  the  plan  I)  M  L  and  C  J  K  as  the 
profiles  of  the  front  and  back  pieces  respectively, 
while  ABE  and  CDF  become  the  miter  lines  or 
elevations  of  the  intersecting  surfaces. 

A  part  of  these  operations  are  carried  out  in 
detail  in  Fig.  339,  where  the  elevation  and  the  plan 
are  drawn  directly  in  line  with  each  other;  the  various 
points  being  represented  by  the  same  letters  in  the 
two  illustrations.  For  the  pattern  of  the  top  piece 
divide  its  profile  A  B  E  by  any  convenient  number  of 
points  (1,  2,  3,  etc.),  from  which  drop  lines  ver- 


tically cutting  the  two  miter  lines  D'  M  and  C'  J  of 
the  plan,  as  shown  (the  figures  of  the  plan  2  to  11 
have  no  reference  to  this  part  of  the  operation).  Upon 
R  8,  drawn  at  right  angles  to  the  direction  of  the  mold, 
lay  off  the  stretchout  of  ABE,  through  which  draw 
the  usual  measuring  lines.  With  the  T-square  placed 
parallel  to  R  8  and  brought  against  the  several  points 
in  the  two  miter  lines  cut  lines  of  corresponding  num- 
ber; lines  traced  through  the  points  of  intersection,  as 
shown  by  R  T  and  U  S,  will  give  the  pattern  of  the  top 
piece.  It  will  be  noticed  that  owing  to  the  contrary 
relation  of  the  two  curves  it  is  necessary  to  have  the 
points  of  the  profile  occur  more  frequently  near  B 
than  E,  as  otherwise  they  would  intersect  the  miter 


Fig.  SS8.— Perspective  View  of  a  Pipe  Describing  a 
Twist  or  Compound  Curve. 


line  D'  M  too  far  apart  near  D',  while  they  would 
occur  more  frequently  than  is  necessary  near  M.  As 
there  is  no  curve  from  A  to  B  of  the  profile,  that  part 
of  the  pattern  from  R  to  S  will  be  a  duplicate  of  the 
plan  view,  consequently  the  curve  from  R  to  the 
measuring  line  drawn  from  S  may  be  traced  from  the 
plan.  The  development  of  the  pattern  for  the  lower 


/'a  ttcni    Problems. 


HI 


Mile  of  the  pipe  is  not  given,  but  it  would  be  accom- 
plished in  exactly  the  same  manner  as  that  of  the  top 
piece,  using  C  D  F  as  the  profile  instead  of  A  B  E. 

For  the   pattern  of  the  front  piece  of  the  pipe, 
divide  its  profile  L  M  D'  by  any  convenient  number 


the  T-square  placed  parallel  to  Q  1  and  brought  against 
the  various  points'  in  B  E  and  D  F  cut  corresponding 
measuring  lines.  Lines  traced  through  the  points  <>f 
intersection,  as  shown  by  Q  P  and  0  N,  will  give  the 
required  pattern. 


/ 

-— 


512  3  t  5  6  1 


Pig.  iS9.— Patterns  for  a  Pipe  Describing  a  Compound  Curve. 


of  points  1,  2,  3,  4,  etc.,  from  which  drop  lines  ver- 
tically, cutting  the  two  miter  lines  1)  F  and  B  E,  as 
shown.  Upon  Q  1,  drawn  at  right  angles  to  the  direc- 
tion of  the  mold,  place  the  stretchout  of  L  M  D', 


The  pattern  of  the  back  piece  not  given  in  the 
illustration  can  be  developed  in  exactly  the  same  man- 
ner as  that  of  the  front  by  using  C'  ,T  K  as  the  profile 
and  proceeding  otherwise  the  same  as  in  the  fore- 


through  which  draw  the  usual  measuring  lines.     With   '  going. 


T/ic    yew    Metal     IFcr/ar    Patient    Jlwk. 


PROBLEM  49. 


The  Construction  of  a  Volute  for  a  Capital. 


It  is  sometimes  desirable  in  designing  capitals  of 
large  size  to  construct  the  volutes  of  the  same  of  strips 
of  metal  cut  and  soldered  together.  The  principal 
characteristic  entering  into  the  design  of  the  volute, 
and  that  which  distinguishes  it  from  an  ordinary  scroll, 
consists  in  a  pulling  out  or  raising  up  of  each  succes- 
sive revolution  of  the  scroll  beyond  the  former,  thus 
producing  a  ram's  horn  effect.  This  feature  of  its  de- 
sign is  also  frequently  embodied  in  the  construction  of 
scrolls  used  to  finish  the  sides  of  large  brackets  or  head 
blocks,  such  as  may  be  seen  by  reference  to  Fig.  87  on 
page  12.  As  all  volutes,  except  those  of  the  Ionic 
order,  always  occur  under  the  corners  of  the  abacus 
and  project  diagonally  from  the  bell  of  the  capital, 
their  forms  can  only  be  correctly  delineated  in  a 
diagonal  elevation. 

In  Fig.  330  is  shown  a  diagonal  elevation  of  a 
portion  of  the  bell  and  abacus  of  a  capital  with  the 
volute.  Immediately  below  the  same,  D  A  C  B  shows 
one-quarter  of  the  plan  of  the  capital,  turned  to  corre- 
spond with  the  elevation,  in  which  the  various  curves 
of  the  volute  have  been  carefully  projected  from  the 
elevation,  as  shown  by  the  dotted  lines.  As  the  pat- 
tern cutter  is  dependent  upon  the  drawing  of  the  plan 
for  his  miter  lines,  considerable  care  must  be  given  to 
this  part  of  the  work.  On  account  of  the  small  scale 
necessary  in  drawing  Fig.  330,  an  enlarged  view  of  the 
plan  of  the  helix  of  the  volute,  as  seen  from  below,  is 
shown  in  Fig.  331,  in  which  the  various  curves  can  be 
followed  throughout  their  course. 

The  volute  as  here  given  consists  of  two  side 
pieces  or  scrolls,  an  outside  cover  or  face  strip,  an  in- 
side cover  and  two  narrow  strips  to  rill  the  space  where 
the  second  curve  of  the  scroll  projects  beyond  the  first. 
The  outside  cover  or  face  strip  extends  from  F  of  tin- 
elevation  to  G,  where  it  is  met  by  the  inside  face  strip, 
which  begins  at  H.  To  obtain  the  pattern  for  the  in- 
side cover,  divide  the  profile  from  II  to  G  into  any 
convenient  number  of  equal  spaces,  and  lay  off  a 
stretchout  of  the  same  upon  the  center  line  of  the 
volute  in  plan,  A  B,  extended  toward  K,  as  shown  bv 


the  nine  spaces  on  the  upper  side  of  the  line.  Drop 
lines  vertically  from  each  of  these  points  intersecting 
the  upper  line  of  the  side  of  the  scroll  in  plan.  Place 
the  J-square  parallel  to  the  stretchout  line  B  K,  and 
bringing  it  successively  against  the  points  in  the  plan, 
drop  lines  cutting  corresponding  lines  of  the  stretch- 
out. Then  a  line  traced  through  the  points  of  inter- 
section, as  shown  from  I  to  J,  will  give  the  shape  of 
the  side  of  the  strip  to  cover  the  space  between  the 
points  H  and  G  of  the  elevation.  A  similar  course  is 
to  be  pursued  in  obtaining  the  outside  cover  or  strip 
extending  from  F  to  G.  This  stretchout  consists  of 
fourteen  spaces,  and  is  shown  on  the  lower  side  of  the 
center  line  A  K,  the  pattern  being  shown  from  L  to 
M.  The  pattern  for  the  remaining  strip  consists  of  a 
stretchout  of  seven  pieces  taken  from  the  profile  be- 
tween G  and  the  termination  of  the  scroll  line.  Points 
from  this  part  of  the  profile  are  intersected  with  two 
miter  lines  in  the  plan,  one  forming  the  outer  line  of 
the  strip,  or  its  finish  against  the  more  projecting  part 
of  the  scroll,  and  the  other  forming  its  finish  against 
the  lower  scroll  or  inner  edge  of  its  first  or  outer  curve. 
In  Fig.  331  the  lines  showing  the  projection  of  the 
inner  part  of  the  volute  beyond  the  outer  curve  are 
clearly  seen.  In  the  lower  half  the  lines  correspond- 
ing to  the  points  1  to  7  of  the  profile  are  shown  by  cor- 
responding numbers.  Lines  dropped  from  the  points  on 
both  these  lines  to  corresponding  lines  of  the  stretchout 
will  give  the  pattern  as  shown  from  M  to  N. 

By  inspection  of  the  drawing  it  will  be  seen  that 
the  outline  of  the  volute,  as  given  in  the  elevation, 
does  not  represent  exactly  the  "true  face"  of  the 
scroll.  As  the  variations  in  the  angle  of  the  side  of  the 
central  part  or  helix  of  the  scroll  are  only  such  as  can  be 
produced  by  the  springing  of  the  metal  necessary  to 
bring  it  into  shape,  no  allowance  need  be  made  for  such 
variation  in  cutting  the  pattern  directly  from  the  eleva- 
tion. Careful  measurements  of  the  stern  or  lower 
part  of  the  volute,  as  shown  in  the  plan,  however,  show 
that,  the  distances  from  point  9  to  points  a  and  i,  if  laid 
olf  on  a  line  parallel  to  A  B,  would  reach  to  points  a' 


Pattern 


143 


and  b\  These  points  projected  back  into  the  elevation 
locate  them  in  that  view  at  a  and  V.  Therefore  the 
outline  of  the  back  of  the  stem  will  have  to  be  ex- 
tended as  shown  by  the  dotted  line  from  F  to  a'.  This 


and  need  not  be  repeated  here.  The  correct  outline,  from 
G  to  b'  is  omitted  to  avoid  confusion  with  the  figures. 

To  avoid  confusion  of  lines  in  dropping  the  points 
from  the  different  parts  of  the  profile  to  the  miter  liues 


Fig.  S3 1,— Enlarged,  View  of  Helity 


PATTERNS 
Fig.  330.— The  Construction  of  a  Volute  for  a  Capital. 


outline  can  be  accurately  obtained,  if  deemed  necessary, 
by  the  raking  process  described  in  connection  with  a 
number  of  other  problems  in  this  section  of  this  chapter, 


and  thence  to  the  stretchout,  only  the  first  and  last  of 
each  series  or  stretchout  have  been  shown  by  dotted 
lines  in  the  drawing. 


144 


Tin-    Sen-    Miiul    Wvrhr    I'ntim,    /;<„,/,; 


PROBLEM   50. 
The  Pattern  for  a  Pyramidal  Flange  to  Fit  Against  the  Sides  of  a  Round  Pipe  Which  Passes  Through  Its  Apex. 


A  pictorial  illustration  of  the  flange  fitting  against 
the  sides  of  the  pipe,  as  stated  above,  is  shown  in  Fig. 
332.  In  Fig.  333  K  L  M  represents  the  elevation  of 


Fig.  333.— Pattern  for  a  Pyramidal  Flange  to  Fit  Against  a 
Round  Pipe. 


pyramid,  and  P  K  T  S  elevation  of  the  pipe  that  is  to 
pass  through  it,  A  B  D  C  being  plan  of  pipe  and  pyra- 
mid. As  the  pyramid  has  four  sides,  each  side  will 
miter  or  fit  against  one-quarter  of  the  profile  oj  the 
pipe,  as  will  be  seen  by  reference  to  the  plan.  Again, 
as  each  side  consists  of  two  symmetrical  halves,  as 
shown  by  the  dotted  line  dividing  the  side  B  D,  one- 
eighth  of  the  profile  of  the  pipe  (as  G  I)  is  all  that  need 


be   used   in   obtaining  the  pattern.      Therefore,  divide 
G  I  into  any  C&nvenient  number  of  parts  and  carry  ver- 
tical lines  to  L  M,  which    represents   one   side  of   the 
pyramid,  and  then,  from  these  points  and 
the   points   L  and  M    carry  lines   at  right 
angles  to  it  indefinitely,  as  shown.      L  M 
in   the  elevation  represents  the  complete 
length  of  one  side  of  the  pyramid,  as   it 
would  be  if  not  cut  by  the  pipe.      Lav  olT 
on  the  line  from  M  the  length  of  one  side 
of  the  base  of  the  pyramid,  as  B  D  in  the 
plan,  as   shown   by  M  M1.      Biseet  M  M1 
at  F,  from  which  point  draw  F  E  parallel 
to  L  M,  cutting   the   line  from  point  L  at 
E.     The  lines  from  K   to  the  points   M 
and   M1   would   give   the   pattern   of    one   side  of    the 
pyramid    if   it  were   not   to   be   cut   by   the  pipe.      It 
simply  remains  now  to  measure  the  width   of  the   pat 
tern  at  the  various  points  of  the  curved  portion,  which 


Fig.  332.— Perspective  View  of  Pyramidal  Flange. 

can  be  done  by  measuring  the  distance  of  each  point  in 
the  profile  G  I,  from  the  center  line  of  the  side  B  D  in 
plan,  and  setting  off  these  distances  upon  lines  of 
corresponding  number  drawn  through  the  pattern  from 
the  line  L  M,  measuring  each  time  from  the  center  line 
E  F.  Thus  the  distance  of  point  4  from  the  center 
line  in  plan  is  set  off  from  the  center  line  of  the  pat- 
tern each  way  upon  lino  4,  and  coincides  with  this 
point  as  previously  established  by  the  lines  drawn  from 
E  to  M  and  M'.  The  distance  of  the  point  3  frcm  the 
center  line  in  the  plan  is  set  off  from  the  center  line  of 
pattern  each  wavMipon  line  3  of  the  pattern.  Point  '1 
is  established  in  the  same  manner.  A  line  traced 
through  the  points  4321234  completes  the  pat- 
tern. 


145 


PROBLEM    51. 

The  Patterns  for  a  Square  Pyramid  to  Fit  Against  the  Sides  of  an  Elliptical  Pipe  Which  Pisses  Through  Its  Center. 


Fa  Fig.  :>:»-i,  A  B  C  D  shows  the  plan  of  a  square 
pyramid,  whose  ;i]ic.\,  if  completed,  would  be  at  E. 
F  II  1  J  shows  the  horizontal  section  of  an  elliptical 
pipe,  against,  the  sides  of  which  the  sides  of  the  pyra- 
mid are  required  to  be  fitted.  From  the  side  A  B  (or 


patterns  will  be  necessary,  one  for  A  B  G  S  to  fit 
against  the  broad  side  of  the  pipe,  and  another  for 
B  G  T  C  to  iit  against  what  might  be  termed  the 
edge  or  narrow  side  of  the  pipe.-  To  obtain  the 
pattern  of  the  side  of  the  pyramid  shown  by  B  G  T1  C 


D  C  C1 

Fig.  SS4.— The  Pattern!  for  a  Square  Pyramid  to  Fit  Against  the  Hides  of  an  Elliptical  Pipe. 


D  C)of  the  plan  is  projected  a  front  elevation,  in  which 
K  L  M  N  shows  the  broad  side  of  the  pipe.  To  the 
right  another  or  side  elevation  is  projected,  in  which 
the  narrow  view  of  the  pipe  is  shown  by  0  P  Q  li. 
An  inspection  of  the  plan  will  show  at  once  that  two 


(or  B  E  C,  if  the  pyramid  were  complete),  first 
divide  that  portion  of  the  profile  of  the  pipe  from 
G  to  H  by  any  convenient  number  of  points,  as  shown 
by  the  small  figures,  from  which,  together  with  B  and 
E,  project  lines  vertically  to  the  elevation  above,  cut- 


146 


The  New  Metal    Worker    I'utl'-rn    lion!.-. 


ting  that  side  in  profile  as  shown  from  E'  to  B1.  At 
right  angles  to  E'  B1  carry  lines  from  each  of  the  points 
indefinitely,  as  shown.  At  any  convenient  distance 
away,  cut  these  lines  by  any  line,  as  E3  V,  drawn 
parallel  to  E1  B'.  Upon  each  of  the  lines  drawn  from 
the  points  in  E1  B',  set  off  from  E3  V3  the  distances 
upon  lines  of  corresponding  number  in  the  plan  meas- 
ured from  E  V.  Thus  upon  line  5  of  the  pattern  set 
off  either  way  from  its  intersection  with  the  line  E1  V3 
a  length  equal  to  the  distance  of  point  5  of  the  plan 
from  the  line  E  V.  Upon  line  4  of  the  pattern  set  off 
distances  equal  to  that  of  point  4  from  E  V  of  the  plan, 
ct.-.  Also  make  V3  C3  and  Vs  B3  equal  to  V  C  and 
V  B  of  the  plan.  Lines  drawn  from  C3  and  B3  toward 
E3  will  meet  the  points  previously  set  off  on  line  5  of 
the  pattern,  indicated  by  T"  and  G1,  and  will  constitute 
the  sides  or  hips  of  the  pattern,  and  a  line  traced 
through  the  points  set  off  on  lines  1  and  5  inclusive 
will  give  the  shape  of  that  portion  of  the  pattern  to  fit 
against  the  pipe.' 

An  exactly  similar  course  is  to  be  pursued   in  ob- 
taining the  pattern  of  the  side  of  the  pyramid  A  S  G  B 


(or  A  K  B  of  tin1  complete  pyramid'),  whose  profile  is 
shown  bv  B3  EJ  of  the  side  elevation,  showing  the  nar- 

J 

row  view  of  the  pipe.  The  pattern  is  shown  at  A4  B" 
G2  S*,  and  the  operation  is  clearly  indicated  by  the 
lines  of  projection. 

If  it  is  desired  to  complete  the  elevations  bv  show- 
ing the  lines  of  intersection  of  the  sides  of  the  pipe 
with  the  sides  of  the  pyramid  shown  respectively  in 
each  elevation,  as  from  c  to  b  and  <-.  toy',  it  can  be  ac- 
complished as  follows  :  To  obtain  the  line  c  i,  erect  lines 
vertically  from  points  ti.  T  and  S  (not  shown),  passing 
through  the  space  between  c  and  b  in  the  front  elevation, 
upon  each  of  which  set  off  the  hight  of  each  point  as 
measured  upon  lines  of  corresponding  number  from  B2 
C"  to  B3  E3,  as  shown  from  R  toward  <?,  in  the  side  eleva- 
tion ;  then  a  line  traced  through  the  points  thus  obtained 
will  give  the  line  ci.  In  the  same  manner  lines  from  tin- 
points  1,  2,  3  and  4  can  be  carried  at  right  angles  to  BaCa 
into  the  side  elevation,  upon  which  to  set  off  hights  of 
corresponding  points  as  measured  from  A'  15'  to  B1  E1, 
as  shown  between  <i  and  1>  in  the  front,  elevation;  then 
a  line  traced  through  the  points  will  give  the  line  ef. 


PROBLEM    52. 


The  Patterns  for  a  Rectangular  Pipe  Intersecting  a  Cylinder  Obliquely. 


In  Fig.  335,  let  A  B  C  represent  the  plan  of  a 
drum  or  cylinder,  and  B  E  D  C  the  plan  of  rectangular 
pipe,  the  profile  of  which  is  shown  by  F  G  II  I.  In 
the  elevation,  J  K  L  M  represents  the  drum,  N  O  P  Q 
the  rectangular  pipe,  and  K  n  O  the  angle  at  which 
they  are  to  intersect.  Draw  the  end  view,  or  plan,  of 
circular  drum  in  line  with  the  elevation,  as  shown. 
Also  extend  0  n  and  P  q  so  a  line  dropped  from  point 
C  of  plan  will  cut  them,  as  shown  by  N  and  Q.  Then 
n  N"  Q  q  is  the  joint  between  the  drum  and  pipe,  as 
shown  in  elevation.  For  the  pattern  of  rectangular 
pipe  proceed  as  follows:  Divide  B  C  of  plan  into  anv 
convenient  number  of  equal  parts,  and  from  these 
points  carry  lines  horizontally  cutting  E  D.  Also  from 
the  points  in  C  B  drop  lines  vertically  cutting  Q  P  and 
N  O.  On  0  P  extended  lay  off  a  stretchout  of  profile 
F  G  II  I,  as  shown  by  1  1'.  transferring  the  spaces  in 
E  D  to  H  G  and  F  I',  and  through  the  points  in  it  draw 
the  usual  measuring  lines,  as  shown.  Place  the  "T-square 


parallel  with  the  stretchout  line  1 1',  and,  bringing  it 
successively  against  the  points  in  the  miter  lines  N  n 
Q  q,  cut  the  corresponding  measuring  lines,  as  indicated 
by  the  dotted  lines.  Lines  traced  through  the  points 
thus  obtained,  as  indicated  by  ihgfi',  will  give  the 
desired  pattern.  It  will  be  observed  that  1  II  li  i  is  a 
duplicate  of  O  P  Q  N,  and  that  G  F///  is  also  a  dupli- 
cate, only  in  a  reversed  position.  The  points  in  h  >/  of 
pattern  arc  derived  from  Q  y,  as  the  points  in  f  i'  of 
pattern  are  derived  from  N  //.  If  the  size  of  tin-  work- 
is  such  as  to  render  it  inconvenient  to  drop  points 
from  the  elevation  to  the  pattern  by  means  of  the  T- 
square,  the  stretchout  line  I  I'  can  be  drawn  where 
convenient,  the  usual  measuring  lines  erected  and  the 
distances  from  O  P  to  points  in  N  »/  and  (.)  */  transferred 
by  means  of  the  dividers  to  lines  of  similar  number 
drawn  from  the  stretchout  line. 

For  the  pattern  or  shppe  of  opening  in  drum,  pro- 
ceed  as  follows:    On  L  M  extended,  as  R  U,  la}-  off  a 


fit//'  /•;.•    I'rnlilems. 


147 


stretchout  of  BO  of  plan,  and  from  the  points  thusob- 
tainc"!  erect  the  usual  measuring  lines,  as  shown.    Place 


in.tr    lines    of    corresponding    number.       Through    the 
points   thus  obtained   trace   the  lines  V  W  and  YX; 


PLAN 


i 


Fig.  335.— Patterns  for  Rectangular  Pipe  Intersecting  a  Cylinder  Obliquely. 


the  T-square  parallel   with  M  L,  and,  bringing  it  suc- 
cessively against  the  points  N  n  and  Q  q,  cut  measur- 


then  V  W  X  Y  will  be  the  shape  of  the  required  open- 
ing in  the  side  of  the  drum. 


PROBLEM    53. 

The  Pattern  lor  the  Intermediate  Piece  of  a  Double  Elbow  Joining:  Two  Other  Pieces  Not  Lying  in  the  Same  Plane. 


In  Fig.  33fi  is  shown  a  front  and  side  view  of  a 
somewhat  complicated  arrangement  of  elbows  such  as 
sometimes  occurs  when  pipes  have  to  be  carried  around 
beams  or  through  limited  openings.  An  inspection  of 


the  drawing  will  show  that  once  the  correct  angle  of 
the  different  elbows  is  ascertained  the  development  of 
the  miters  will  be  quite  simple,  and  is  the  same  as 
those  occurring  in  several  of  the  problems  preceding 


148 


Tin-   Sf.w  Metal    Worker   Pattern  B<>«k. 


this.  The  lower  section  of  the  pipe  rises  vertically  to 
the  first  elbow,  B,  from  which  it  must  be  carried  up- 
ward a  distance  equal  to  C  M,  to  the  left  a  distance 
equal  to  B  M,  as  shown  in  the  front  view,  and  back  a 
distance  equal  to  o  C,  as  shown  by  the  side  view. 


methods  employed  in  drawing  the  two  views  shown  in 
Fig.  336  will  be  of  assistance  to  the  pattern  cutter. 
According  to  the  principles  of  projection  each  indi- 
vidual point  must  appear  at  the  same  hight  in  both  ele- 
vations, and  at  the  same  distance  right  or  left  and  for- 


2.  s 

FRONT  VIEW  SIDE  VIEW 

Fig.  836.— Elevations  of  Double  Elbow. 


Fig.  a38.— Diagram  Used  in  Obtaining 
Correct  Side  View  of  Upper  Elbow. 


Fig.  337.— Correct  Side  View  of 
Lower  Elbow. 


1    2 


Fig.  840.— Method  of  Obtaining  the  Pattern  of  Middle 
Portion  in  One  Piece. 


Fig.  339.— Correct  Side  View  of  Upper  Elbow. 
The  Pattern  for  the  Intermediate  Piece  of  a  Double  Elbow  Joining  Two  Other  Pieces  Nut  Lying  in  the  Same  Plane. 


From  the  elbow  C  it  then  rises  vertically,  as  seen  in 
front,  but  really  toward  the  observer  as  shown  by  the 
side  view.  The  problem  then  really  consists  in  find- 
ing the  correct  angles  of  the  elbows,  and  becomes  a 
question  of  draftsmanship  rather  than  of  pattern  cut- 
ting. Some  suggestions  then  with  regard  to  the 


ward  or  back,  with  reference  to  the  center  lines  of  the 
plan.  As  front  and  side  views  are  here  required, 
begin  by  first  placing  the  given  plan  in  two  positions, 
turning  those  sides  of  it  to  the  bottom  which  corre- 
spond to  the  sides  required  in  the  elevations,  and 
proceed  by  erecting  the  center  lines  of  the  differ- 


Pattern    Problems. 


149 


ent  pieces  in  their  proper  positions  and  building  the 
pipe  around  them,  so  to  speak.  The  plan  being 
a  circle,  the  different  sides  can  only  lie  indicated  by 
numbering  the  points,  as  will  be  seen  by  referring 
to  the  plans,  point  2  appearing  in  front  in  the  front 
elevation,  and  point  3  appearing  in  front  in  the  side 
elevation.  The  plans  having  been  so  arranged  and 
corresponding  parts  in  both  given  the  same  number, 
proceed  now  to  erect  the  center  line  of  the  lower  sec- 
tion, making  the  bight  of  the  first  bend,  B,  the  same  in 
both  views,  as  indicated  by  the  dotted  horizontal  line. 
From  this  point  the  center  line  is  continued  in  both 
views,  giving  it  its  proper  inclination  to  the  left  in  the 
front  view,  and  to  the  right  in  the  side  view,  all  accord- 
ing to  the  specified  requirements,  thus  establishing  the 
point  C,  making  it  agree  in  bight  in  both  views.  From 
this  point  the  pipe  appears  inclined  only  in  the  side 
view,  which  means  that  it  leans  toward  the  observer 
in  the  front  view.  Next  draw  the  outlines  of  the  pipe 
at  equal  distances  from  the  center  line  and  on  either 
side  of  it  throughout  the  entire  course  of  the  pipe  in 
both  views,  deriving  them  from  the  points  of  plans  1 
and  3  in  the  front  view  and  2  and  4  in  the  side  view. 
Their  intersection  in  the  front  view  will  give  definitely 
the  positions  in  the  miter  of  points  1',  1",  and  3',  3", 
and  in  the  side  view  of  points  2',  2',  and  4',  4J.  As 
point  3' has  been  established  in  the  front  view,  if  a  line 
be  carried  horizontally  across  till  it  intersects  the  line 
from  point  3  of  the  side  view,  it  will  give  the  hight  of 
point  3'  in  the  miter,  as  shown  in  the  front  view.  In  the 
same  manner  a  horizontal  line  from  1'  in  front,  inter- 
secting the  perpendicular  f?om  point  1  in  plan  of  side, 
will  give  the  true  hight  of  point  1'  in  the  side  view. 
A  careful  inspection  of  the  dotted  lines  of  Fig.  336  will 
make  the  subsequent  operations  necessary  to  the  com- 
pletion of  the  elevations  clear  to  the  reader  without 
further  explanation.  Since  neither  of  the  views  gives 
a  true  side  view  of  the  intermediate  piece,  one  must 
be  constructed  from  the  facts  now  known,  so  as  to  get 
the  true  angle  of  the  elbow  B.  By  dropping  a  vertical 
line  from  the  point  C  of  the  front  view  into  the  plan  it 
will  appear  that  the  horizontal  distance  between  the 
points  C  and  B  would  be  measured  by  the  line  K  P  of 
the  plan;  but  bv  further  reference  to  the  side  eleva- 
tion the  position  of  the  point  C  is  found  to  be  to  the 
right  of  its  center  line  by  a  distance  equal  to  B  C' 
of  the  plan ;  therefore,  if  this  distance  be  set  off  on  the 
vertical  line  from  the  point  E  in  the  plan  below  the 
front  view,  which  is  indicated  by  E  C,  the  point  C  will 
determine  the  true  position  in  the  plan  of  the  point 


C  of  the  elevations,  and  the  distance  C  P  will  be  its 
horizontal  distance  from  B.  Since,  now,  its  vertical 
distance  cap  easily  be  obtained  from  either  front  or  side 
elevation,  a  new  diagram  can  now  be  easily  constructed 
which  shall  contain  the  proper  dimensions  to  obtain  a 
correct  side  view  of  this  elbow.  Proceed,  then,  to 
construct  diagrams  shown  in  Fig.  337,  making  C  M 
equal  to  C  M,  Fig.  336,  M  B  equal  to  C  P  of  the  plan, 
Fig.  336 ;  a  line  connecting  the  points  C  and  B  will 
represent  the  center  line  of  the  intermediate  portion  of 
the  pipe  and  give  its  true  relation  to  the  vertical  por- 
tion whose  center  line  is  represented  by  B  H,  Fig. 
337.  By  drawing  the  outlines  of  the  pipe  at  the  re- 
quired distance  on  either  side  of  the  center  lines  B  H 
and  B  C,  a  correct  side  view  of  the  miter  is  obtained. 
Since,  as  has  been  referred  to  above,  the  upper  portion 
of  the  pipe  appears  vertical  in  one  view  and  inclined  in 
the  other  (see  Fig.  336),  a  correct  side  view  of  the 
upper  elbow  is  more  difficult  to  be  obtained.  While 
different  methods  may  be  devised  for  obtaining  it,  the 
following  is  perhaps  the  simplest :  As  the  upper  sec- 
tion of  the  pipe,  as  shown  by  Fig.  336,  is  of  indefinite 
length,  any  point  may  be  assumed,  as  D,  from  which 
to  take  measurement  for  obtaining  the  angle  of  the 
upper  elbow.  Since  the  true  length  of  the  line  C  B  of 
either  elevation  has  already  been  obtained  and  given 
in  Fig.  337,  and  since  the  true  length  of  the  part  C  D 
can  be  derived  from  the  side  view  of  Fig.  336,  it  is 
necessary  only  to  obtain  the  true  distance  between  the 
points  D  and  B  of  the  elevations  to  obtain  the  proper 
angle  at  the  point  C.  By  dropping  a  vertical  line  from 
the  point  D  to  a  horizontal  line  drawn  from  the  point  C 
in  the  side  view,  Fig.  336,  the  horizontal  distance  be- 
tween C  and  D  may  be  obtained.  By  tranf erring  this 
distance,  o  C,  to  the  plan  of  the  front  view,  and  locat- 
ing its  distance  from  C,  as  indicated  by  D.  this  point 
will  give  the  true  position  of  the  point  D  in  the  plan, 
and  the  line  1)  P  will  give  the  true  horizontal  distance 
between  the  points  D  and  B.  In  Fig.  338  let  the 
distance  0  C  be  equal  to  the  line  D  P  of  Fig.  336.  At 
point  O  erect  a  perpendicular,  O  D,  making  the  dis- 
tance O  D  equal  to  o  D  of  the  side  elevation,  Fig.  336. 
From  the  point  C  drop  a  perpendicular,  C  B,  making 
that  distance  equal  to  the  vertical  hight  between  the 
points  C  and  B,  as  measured  on  line  C  M  of  the  front 
view ;  a  diagonal  line  connecting  the  points  D  and  B 
will  readily  be  seen  to  give  the  true  distance  between 
the  points  bearing  those  letters  in  Fig.  336.  Proceed 
now  to  construct  the  triangle  shown  in  Fig.  339,  mak- 
ing C  B  equal  to  C  B  of  Fig.  337.  From  C  as  a  center, 


150 


Xf/r   .!/<'/"/    Worker  Puttcm    IlooL: 


with  a  radius  equal  to  C  D  as  obtained  from  the  side 
view  in  Fig.  336,  draw  a  small  arc.  which  intersect 
with  the  arc  drawn  from  the  point  R,  with  a  radius 
equal  to  B  D  as  obtained  in  P'ig.  338  ;  this  will  give  the 
correct  angle  of  the  upper  elbow  at  C.  A  complete 
view  of  the  miter  may  be  obtained  by  further  adding 
outlines  of  the  pipe  at  equal  distances  on  either  side  of 
the  center  lines,  and  connecting  their  angles,  as  shown 
by  the  line  bf.  Having  now  obtained  two  correct  side 
views  of  the  two  elbows,  the  problem  of  obtaining  the 
patterns  for  the  same  can  be  solved  by  the  regular 
method. 

To  obtain  the  pattern  for  the  middle  portion  in 
one  piece  further  calculations,  however,  will  be  re- 
quired. This,  of  course,  could  be  obviated  by  making 
a  slip  joint  in  the  middle  portion  of  the  pipe,  by  means 
of  which  the  two  elbows  could  be  made  separate,  and 
then  simply  turned  upon  each  other  till  the  required 
angle  is  obtained.  But  as  it  might  be  desirable  to 
make  the  pattern  of  the  middle  portion  in  one  piece  some 
means  must  be  employed  of  ascertaining  just  how  far 
one  elbow  would  have  to  be  turned  upon  the  other 
were  they  made  separately.  As  the  seam  in  pipe  con- 
taining elbows  is  usually  made  at  either  the  shortest  or 
the  longest  point  of  the  miter,  it  may  be  easily  seen,  by 
an  inspection  of  Fig.  336,  that  a  line  from  the  shortest 
point,  or  throat,  b  of  the  upper  miter  of  the  piece  in 
question,  would  not  meet  the  longest  point,  or  point  a, 
Fig.  337,  in  the  miter  of  the  other  end,  and  some 
means  must  be  devised  for  obtaining  the  real  position 
of  these  points,  of  which  the  following  is  perhaps  the 
simplest :  From  either  of  the  points  D  or  B,  Fig.  339, 
draw  a  line  through  the  point  i,  continuing  it  to  the 
further  side  of  the  triangle,  as  indicated  by  the  line 
B  X.  Lay  off  the  distance  D  X  upon  the  line  D  C  of 
the  side  view,  Fig.  336,  thereby  locating  the  position 
of  the  point  x  in  that  view.  A  line  connecting  this 
point  with  point  B  must  intersect  the  miter  line  2",  4:', 
in  this  view  at  the  same  point  which  it  does  in  Fig. 
339,  thereby  locating  its  position  just  as  much  as 
in  Fig.  339.  This  point  having  been  obtained,  its 
equivalent  upon  the  lower  miter  may  be  found  by 


means  of  a  line  drawn  parallel  to  the  center  line  of  the 
middle  portion,  intersecting  it  at,  the  point  ij,  from 
which  point  it  can  be  carried  vertically  to  the  plan,  as 
shown  by  Z,  where  its  distance  from  other  points  can 
be  measured  with  accuracy.  The  position  of  the  point 
a  in  Fig.  337  will  readily  be  seen  to  be  at  point  //  in 
the  plan  of  the  front  view,  Fig.  336.  By  transferring 
the  point  Z  from  the  plan  of  the  side  view  to  the  plan 
of  the  front  view,  which  can  be  done  by  measuring  its 
distance  from  either  of  the  points  2  or  3,  the  relative 
position  of  the  points  h  and  Z  upon  the  same  circle  will 
be  apparent.  Fig.  340  shows  a  diagram,  in  which  a 
correct  side  view  of  the  two  elbows  is  shown,  giving 
,  the  proper  distance  between  the  points  B  and  C. 
Considering  the  lower  one  to  be  in  its  proper  and  lixed 
position,  the  profile  is  constructed  and  divided  into 
points  for  the  purpose  of  obtaining  a  stretchout  and 
the  miter  pattern  according  to  the  usual  method,  the 
|  stretchout  being  shown  upon  the  line  E  F  in  the 
j  profile  and  point  8  will  readily  be  seen  to  correspond 
with  point  h  in  the  plan  of  the  front  view.  The  posi- 
tion of  the  point  Z  in  the  same  plan  can  be  obtained 
by  measuring  its  distance  from  point  h  and  transferring 
it  to  Fig.  340,  as  indicated  by  M.  As  the  point  I  of 
the  upper  elbow  is  in  relation  to  the  highest  point,  or 
«,  of  the  lower  elbow  as  the  point  M  is  to  the  point  8 
in  the  profile,  it  becomes  necessary  to  place  the  point 
8  in  the  stretchout  of  the  upper  elbow  as  far  from  the 
point  8  on  the  stretchout  of  the  lower  one  as  the  dis- 
tance from  8  to  M  in  the  profile,  which  is  shown  by 
m  in  the  stretchout.  The  stretchout  of  the  upper 
elbow  is  thus  moved,  as  it-were,  in  its  relation  to  the 
stretchout  of  the  lower  elbow,  that  portion  of  it  which 
extends  beyond  the  point  1  at  the  left  end  being  added 
to  the  other  so  as  to  make  the  seam  continuous.  The 
points  are  then  dropped  from  the  profile  to  the  two 
rniter  lines,  and  thence  into  measuring  lines  of  corre- 
sponding number  in  the  stretchout.  Lines  traced 
through  the  points  of  intersection,  as  shown  by  Y  I*  R 
X,  will  be  the  required  pattern.  The  miters  for  the 
upper  and  lower  sections  would,  of  course,  be  inverted 
duplicates  of  the  adjacent  ends  of  the  middle  piece. 


Pattern  ProU<-inx. 

PROBLEM    54. 

A  Joint  Between  Two  Pipes  of  the  Same  Diameter  at  Other  Than  Right  Angles. 


151 


Let  L  F  D  E  K  I  II  _\l  of  Fig.  341  represent  the 
elevation  of  two  pipes  of  the  same  diameter  meeting  at 
the  angle  M  H  I,  for  which  patterns  are  required. 
Draw  the  profile  or  section  A'  B'  C'  in 
line  with  the  branch  pipe,  and  the  section 
A  B  C  in  line  with  the  main  pipe.  As 
both  pipes  are  of  the  same  diameter,  and 
the  end  of  one  piece  comes  against  the 
side  of  the  other  piece,  both  halves  of 
the  branch  pipe  (dividing  at  the  point  B) 
will  miter  with  one-half,  B  D,  of  the  main 
pipe.  By  projecting  lines  through  the 
elevations  of  each  piece  from  the  points  B 
or  4  of  their  respective  profiles  the  point 
G  is  obtained,  which,  being  connected  with 
points  F  and  H,  gives  the  required  miter 
line.  Space  both  the  profiles  into  the 
same  number  of  equal  divisions,  com- 
mencing at  the  same  point  in  each.  For 
the  pattern  of  the  arm  proceed  as  follows  : 
Lay  off  the  stretchout  O  N  opposite  the 
end  of  the  arm  and  draw  the  usual  meas- 
uring lines  at  right  angles  through  it,  as 
shown.  Place  the  T-square  at  right 
angles  with  the  arm,  or,  what  is  the  same, 
parallel  with  the  stretchout  line,  and, 
bringing  the  blade  successively  against  the 
points  in  the  miter  line  F  G  H,  cut  the  corresponding 
measuring  lines.  Through  the  points  thus  obtained 
trace  the  line  PEST,  which  will  form  the  end  of  the 
pattern  required.  For  the  pattern  of  the  main  pipe 
proceed  as  follows :  Opposite  one  end  lay  off  the 
stretchout,  as  shown  by  V  Y,  and  opposite  the  other 
end  lay  off  a  corresponding  line,  as  shown  by  U  X. 
Connect  U  V  and  X  Y.  From  so  many  of  the  points 
in  the  stretchout  line  V  Y  as  represent  points  in  the 
half  of  the  profile  BAD  draw  the  usual  measuring 
lines.  With  the  T-square  placed  parallel  to  the  mold- 
ing D  I,  drop  the  points  from  the  profile  onto  the 
miter  line  F  G  H;  then,  placing  it  at  right  angles  to 
the  molding,  drop  lines  from  the  points  in  the  miter 
line  intersecting  the  corresponding  measuring  lines. 
A  line  traced  through  these  points  of  intersection,  as 
F'  Z  H1  W,  will  describe  the  shape  required.  The 
position  of  the  seam  in  both  the  arm  and  the  main  pipe 
is  determined  by  the  manner  of  numbering  the  spaces 


in  the  stretchout.  In  the  illustration  the  seam  in  the 
arm  is  located  in  the  shortest  part,  or  at  a  point  corre- 
sponding to  1  of  the  profile.  Accordingly,  in  number- 


lY 


Fig.  S41  —A  Joint  between  Two  Pipes  of  the  Same  Diameter  at 
Other  than  Right  Angles. 


ing  the  divisions  of  the  stretchout,  that  number  is 
placed  first.  In  like  manner  the  seam  in  the  main  pipe 
is  located  at  a  point  opposite  the  arm.  Therefore,  in 


152 


The  New  Metal    Worker  Pattern   Book. 


numbering  the  spaces  in  the  stretchout  commence 
at  1,  which,  as  will  be  seen  by  the  profile,  represents 
the  part  named.  If  it  were  desirable  to  make  the  seam 
come  on  the  opposite  side  of  the  main  pipe  from  where 
it  has  been  located — that  is,  come  directly  through  the 
opening  made  to  receive  the  arm — the  numbering  of  the 
stretchout  would  have  been  begun  with  7.  In  that  case 
the  opening  F1  W  IT  Z  would  appear  in  two  halves,  and 
the  shape  of  the  pattern  would  be  as  though  the  pres- 
ent pattern  were  cut  in  two  on  the  line  7  and  the  two 
pieces  were  joined  together  on  lines  1.  By  this  expla- 
nation it  will  be  seen  that  the  seams  may  be  located 


during  the  operation  of  describing  the  pattern  wherever 
desired.  It  is  not  necessary,  as  prescribed  at  the  out- 
set of  this  problem,  that  both  profiles  should  be  spaced 
off  exactly  alike.  Any  set  of  spaces  will  answer  quite 
as  well,  provided  there  be  points  in  each  exactly  half 
way  between  A  and  B  of  either  profile — that  is,  where 
points  4  are  now  located.  They  are  spaced  alike  in  this 
case  to  show  that  lines  dropped  from  points  of  the  same 
number  in  each  profile  arrive  at  the  same  point  on  the 
miter  line,  and  that  therefore  when  both  pipes  are  the 
same  diameter  and  their  axes  intersect,  one  profile  may 
be  used  for  the  entire  operation. 


910 


9-10 


Note. — In  the  nineteen  problems  immediately  following,  the  conditions  are  such  that  it  will  be 
necessary  to  obtain  the  miter  line  from  the  data  given  by  the  operation  of  raking  before  the  straight- 
forward work  of  laying  out  the  patterns  can  be  begun.  However,  as  certain  parts  of  the  work  of 
raking  the  miter  line  and  of  laying  out  the  pattern  are  common  to  both  operations,  the  two  are  usually 
carried  along  together,  and  therefore  such  points  and  spices  should  be  assumed  upon  the  profiles  at  the  out- 
set as  will  be  required  in  the  final  stretchout. 

PROBLEM    55. 
To  Obtain  the  Miter  Line  and  Pattern  for  a  Straight  Molding  Meeting  a  Curved  Molding  of  Same  Profile. 


In  Fig.  342,  let  F  G  J  K  represent  a  piece  of 
straight  molding  joining  a  curved  mold,  G  H  I  J, 
the  profiles  of  the  straight  and  curved  molding  being 
the  same.  To  obtain  the  miter  line  or  line  of  joint, 
Gr  J,  proceed  as  follows :  Draw  the  profile  in  line  with 
the  straight  molding,  as  shown  by  C  D  E,  and  divide 
into  any  convenient  number  of  parts.  From  the  divi- 
sions in  the  profile  draw  lines  parallel  to  F  G  in  the 
direction  of  the  miter  indefinitely,  and  also  in  the 
opposite  direction,  cutting  the  vertical  line  C  E  of  the 
profile,  as  shown  by  the  small  figures,  which  corre- 
spond in  number  to  the  divisions  on  the  profile.  From 
B,  the  center  from  which  the  curved  molding  is  struck, 
draw  the  line  B  A  through  the  molding,  as  shown. 
Transfer  the  hights  of  the  various  points  of  the  profile 
as  obtained  on  the  line  C  E  to  the  line  A  B,  placing  the 
point  E  at  the  point  o  of  the  intersection  of  the  lower 
line  of  the  curved  molding  with  the  line  A  B,  all  as 
shown  by  X  o.  Then,  with  B  as  a  center,  draw  arcs 
from  the  divisions  on  the  line  X  o,  intersecting  lines 
of  corresponding  numbers  drawn  from  the  profile 
parallel  to  the  lines  of  the  straight  molding.  A  line 
traced  through  these  intersections,  as  shown  by  Gr  J, 
will  be  the  required  miter  line,  and,  as  will  be  seen,  is 
not  a  straight  line.  To  obtain  the  pattern  for  the 


K/ 

I 

| 

:  N 

i 

! 

1 

-fi\ 

! 

ii 
n 

\ 

n 

\ 

1 
I 

ij 

I 

, 

\ 

i 

•J 

! 

i 

V 

i 

N 


PATTERN 


Fig.    Sj£.—The    Miter   Line   between   a    Curved   and   a   Straight    Molding 
of  the  Same  Profile. 


Pattern    /'/-otilems. 


153 


straight  molding,  draw  the  line  L  N  at  right  angles  to 
it,  upon  which  place  the  stretchout  of  the  profile  C  D 
K,  as  shown  by  the  small  ligures.  At  right  angles  to 
the  stretchout  line  L  X.  and  through  the  points  in  it, 
draw  the  usual  measuring  lines.  With  the  T-square 
placed  at  right  angles  to  K  J,  bring  it  successively  against 


the  points  forming  the  miter  line  G  ,],  and  cut  lines  of 
corresponding  number  in  the  stretchout.  Then  a  line 
traced  through  these  points  of  intersection  will  form  the 
miter  end  of  the  pattern  shown  by  L  M  N  O.  The 
methods  employed  in  obtaining  the  patterns  for  the 
curved  portions  are  treated  in  Section  2  of  this  chapter. 


PROBLEM    56. 

A  T-Joint  Between  Pipes  of  Different  Diameters. 


In  Fig.  343  it  is  required  to  make  a  joint  at  right 
angles  between  tin  smaller  pipe  D  F  G  E  and  the  larger 


used  in  both  operations  the  following  course  will  be 
most  economical. 

At  a  convenient  distance  from  the  end  of  the 
smaller  pipe  in  each  view  draw  a  section  of  it.     Space 
these  sections  into  any  suitable  number  of  equal  parts, 
commencing  at  corresponding  points  in  each,  and  set- 
ting off  the  same  number  of  spaces,  all  as  shown  by  A 
B  C  and  A1   B1   C1.      From   the 
points  in  A  B  C  draw  lines  down- 
ward   through    the   body    of    the 
large  pipe  indefinitely.     From  the 
points  in  A1  B'  C'  drop  point  onto 
the  profile  of  the   large  pipe,  as 
shown  by  the  dotted  lines.      For 
the  pattern  of  the  smaller  pipe  the 
requirements  are  its  profile  A  B' 
C1    and    the    line    F1    G',    which 


-  b 

-  6 

-  -7 

-  e 

-  9 


Fiij.  3(3.— A  1-Joint  between  Pipes  of 
Different  Diameters. 

pipe  H  K  L  I.  For  this  purpose  both  a  side  and  an 
end  view  are  necessary.  As  the  two  pieces  forming  the 
J  are  of  different  sections  this  problem  really  consists 
of  two  separate  operations,  but  as  certain  steps  can  be 


is  the  outline  of  the  surface  against  which  it  miters, 
and  therefore  its  miter  line.  Therefore,  take  the 
stretchout  of  A'  B1  C1  and  lay  it  off  at  right  angles 
opposite  the  end  of  the  pipe,  as  shown  by  V  W.  Draw 
the  measuring  lines,  as  shown.  Then,  with  the  T-square 
set  parallel  to  the  stretchout  line,  and  brought  succes- 
sively against  the  points  between  F'  and  G'  upon  the 
profile  of  the  large  pipe,  cut  corresponding  measuring 
lines,  as  shown.  Then  a  line  traced  through  these 
points,  as  shown  from  X  to  Y,  will  form  the  end  of  the 
pattern. 

For  the  pattern  of  the  larger  pipe  the  stretchout  is 
taken  from  the  profile  view  F1  G'  L'  and  laid  off  at 
right  angles  to  the  pipe  opposite  one  end,  as  shown  by 
"N  P.  A  corresponding  line,  M  O,  is  drawn  opposite 
the  other  end,  and  the  connecting  lines  M  N  and  0  P 
are. drawn,  thus  completing  the  boundary  of  the  piece 
through  which  an  opening  must  be  cut  to  meet  or 
miter  with  the  end  of  the  smaller  pipe.  According  to 
the  rule  given  in  Chapter  V,  a  profile  and  a  rniter  line 
are  necessary.  The  profile  F'  G1  L'  has  already  been 
stated,  but  no  line  has  yet  been  drawn  in  the  elevation 


154 


The  Neiv  Metal    \Vorker  Pattern   Book. 


of  the  larger  pipe  which  shows  its  connection  with  the 
smaller  pipe.  This  can  only  be  found  l>v  projecting 
line's  from  the  points  dropped  upon  K'  (}'  through  the 
elevation  till  they  intersect  with  lines  previously  drawn 
from  the  'profile  A  B  C,  as  shown  between  F  and  G. 
F  G  then  constitutes  the  miter  line.  For  economy's 
sake,  then,  the  spaces  1  to  4  previously  obtained  in  the 
profile  are  duplicated  upon  the  stretchout,  as  shown,  to 
which  are  added  as  many  more  (4  to  10)  as  are  neces- 
sary. As  the  points  1  to  4  have  already  been  dropped 
upon  the  miter  line  in  its  development  it  is  now  only 
necessary  to  drop  them  parallel  to  the  stretchout  line 


into  measuring  lines  of  corresponding  number,  when  a 
line  traced  through  the  points  of  intersection,  as  shown 
by  R  S  T  U,  will  give  the  pattern  of  the  opening 
required. 

It  may  be  noticed  that  the  development  of  the 
miter  line  F  G  is  not  really  necessary  in  this  case,  as 
the  points  are  really  dropped  from  the  profile  ABC 
right,  through  the  elevation  till  they  intersect  the 
measuring  lines.  This  happens  in  consequence  of  the 
arm  or  smaller  pipe  being  at  right  angles  to  the  larger 
one.  Different  conditions  are  shown  in  Problems  57 
and  58  following. 


PROBLEM    57. 
The  Joint  Between  Two  Pipes  of  Different  Diameters  Intersecting  at  Other  Than  Right  Angles. 


Let  ABC,  Fig.  344,  be  the  size  of  the  smaller 
pipe,  and  Y  N1  Z  the  size  of  the  larger  pipe,  and  let 
II  L  M  be  the  angle  at  which  they  are  to  meet.  Draw 
an  elevation  of  the  pipes,  as  shown  by  G  K 1  0  N  M  L  H, 
placing  the  profile  of  the  smaller  pipe  above  and  in  line 
with  it,  as  shown,  also  placing  a  profile  of  the  larger 
pipe  in  line  with  its  elevation,  as  shown.  In  this 
problem  the  profiles  of  the  moldings  or  pipes  are  given, 
but  the  line  representing  their  junction  must  be  ob- 
tained before  going  ahead. 

To  obtain  this  miter  line,  first  place  a  duplicate  of 
the  profile  of  the  smaller  pipe  in  position  above  the  end 
view  of  the  larger  pipe,  as  shown  by  A1  B'  C1,  the  cen- 
ters of  both  being  on  the  same  vertical  line,  C'  N1. 
Divide  both  profiles  of  the  small  pipe  into  the  same 
number  of  spaces,  commencing  at  the  same  point  in  each. 
From  the  points  in  A  B'  C  project  lines  indefinitely 
through  the  elevation  of  the  arm,  as  shown.  From  the 
points  in  A'  B'  C1  drop  lines  on  to  the  profile  of  the 
large  pipe,  and.  from  the  points  there  obtained  carry  lines 
across  to  the  left,  producing  them  until  they  intersect 
corresponding  lines  in  the  elevation.  A  line  traced 
through  these  several  points  of  intersection  gives  the 
miter  line  K  L,  from  which  the  points  in  the  two 
patterns  are  to  be  obtained.  For  the  pattern  of  the 
small  pipe  proceed  as  follows  :  Opposite  the  end  lay  off 
a  stretchout,  at  right  angles  to  it,  as  shown  by  E  F. 
Through  the  points  in  it  draw  the  usual  measuring 
lines,  as  shown.  In  the  developing  of  the  line  K  L 
the  points  have  already  been  dropped  upon  the  miter 
line.  It  therefore  only  remains  to  carry  them  into  the 
stretchout,  which  is  done  by  placing  the  -square  at 


1  1    1     1      1 
II       II 
II       1 

H  |           i        1 

i|  i           ii 
i|  i     1      i        | 

Y6;( 

Fig.  344.— The  Joint  between  Two  Pipes  of  Different  Diameters 
Intersecting  at  Other  than  Right  Angles. 


Pattern  Problems. 


155 


right  angles  with  the  pipe,  and,  bringing  it  successively 
against  tin1  ]iDints  iii  the  miter  line  K  L,  cut  the  corre- 
sponding measuring  lines,  as  shown  liy  the  dotted 
lines.  A  line  traced  through  the  points  thus  obtained 
will  give  the  pattern  of  the  end  of  the  ami,  as  indi- 
cated. 

For  the  pattern  of  the  large  pipe  proceed  as  fol- 
lows: Opposite  one  end,  and  at  right  angles  to  it,  lay 
off  a  stretchout  line,  as  shown  bv  K  S  In  spacing  off 
this  stretchout  u  is  l>est  to  transfer  the  spaces  from  4 
to  4  as  they  exist,  as  by  so  doing  measuring  lines  will 
result  which  will  correspond  with  points  already  exist- 
ing in  the  miter  line  K  L,  thereby  saving  labor,  as  in 


the  case  of  the  smaller  pipe,  and  also  avoiding  con- 
fusion. The  other  points  in  the  profile  arc  taken  at 
convenience,  simply  for  stretchout  purposes.  Draw  a 
corresponding  line,  P  T,  opposite  the  other  end,  and 
connect  P  R  and  T  S.  In  laying  off  the  stretchout 
R  S,  that  number  is  placed  first  which  represents  the 
point  at  which  it  is  desired  the  seam  shall  come.  For 
the  shape  of  the  opening  in  the  pattern,  draw  measur- 
ing lines  from  the  points  4,  3,  2,  1,  2,  3,  4,  as  shown,  and 
intersect  them  by  lines  dropped  from  corresponding 
points  in  the  miter  line.  Through  the  points  thus  ob- 
tained trace  the  line  U  V  W  X,  which  will  represent 
the  shape  of  the  opening  required. 


PROBLEM    58. 

The  Joint  Between  an  Elliptical  Pipe  and  a  Round  Pipe  of  Larger  Diameter  at  Other  Than 

Right  Angles.— Two  Cases. 


In  Fig.  345  J  K  L  M  is  the  side  elevation  of  the 
round  pipe  and  E  F  G  II  that  of  the  elliptical  pipe 
joining  the  larger  pipe  at  the  angle  F  G  J.  In  the 


WJ 


ioQ 


Fig.  345  —The,  Joint  between  an  Elliptical  Pipe  and  a  Round  Pipe  of  Larger 
Diameter  at  Other  than  Right  Angles.  First  Case.  The  Major  Axis  of  the 
Elliptical  Pipe  Crossing  the  Round  Pipe. 

end  elevation  T  S  I  shows  the  profile  of  the  round 
pipe  and  U  R  S  T   the  intersection  of  the   elliptical 


pipe  whose  profile  is  shown  at  A  B  C  D  and  N  0  P  Q, 
respectively,  in  the  side  and  end  views.  From  an  in- 
spection of  the  drawings  it  will  be  seen  that  the  side 
elevation  shows  the  narrow  view  of  the  elliptical  pipe, 
while  the  end  elevation  shows  its  broad  view,  or  in 
other  words,  that  the  profile  of  the  elliptical  pipe  is  so 
placed  that  its  major  axis  crosses  the  round  or  larger 
pipe.  In  Fig.  346  the  elevations  show 
the  same  pipes  intersecting  at  the  same 
angle,,  but  with  the  difference  that  the 
profile  of  the  elliptical  pipe  is  so  placed 
that  its  minor  axis  crosses  the  round 
pipe.  The  reference  letters  and  figures 
are  the  same  in  the  two  drawings  and 
the  following  demonstration  will  apply 
equally  well  to  either: 

By  way  of  getting  ready  to  lay  out 
the  miter,  it  will  first  be  neeessarv  to 
obtain  a  correct  elevation  of  the  miter 
line  or  intersection  between  the  two 
pipes,  as  sliown,  from  H  to  G.  To  do 
this  divide  the  two  profiles  A  B  C  D 
and  N  O  P  Q  into  the  same  number  of 
equal  parts,  commencing  at  the  same 
points  in  each.  Draw  lines  from  the 
points  in  N  O  P  Q,  parallel  with  U 
T,  cutting  T  S.  In  a  similar  manner  draw  lines  in- 
definitely from  the  points  in  A  B  C  D,  parallel  with 


156 


The  Nno   Mni<il    Worker   Pattern    Hook. 


II  E,  as  shown.  From  the  points  in  T  S  draw  linos 
parallel  with  M  .],  which  produce  until  corresponding 
lines  from  the  two  profiles  intersect.  Through  the 
several  points  of  intersection  thus  obtained  draw  the 


right,  angles  with  H  E,  or  parallel  with  V  W,  and 
brought  successively  against  the  points  in  the  miter 
line  II  G,  cut  corresponding  measuring  lines.  A  line 
traced  through  the  points  thus  obtained,  as  shown  by 


Fig.  $46.— Second  Case. 


The  Minor  Axis  of  the  Elliptical  pipe  Crossing  the 
Round  Pipe. 


miter  line  H  G.  For  the  pattern  of  E  F  G  II  proceed 
as  follows :  On  E  F  extended,  as  V  W,  lay  off  a 
stretchout  of  profile  A  B  C  I),  through  the  points  in 
which  draw  the  usual  measuring  lines  at  right  angles 
to  the  stretchout  line.  With  the  T-square  placed  at 


X  Y  Z,  will  give  the  miter  cut  required,  and  V  "W  X  Y 
Z  shows  the  entire  pattern. 

The  method  of  obtaining  the  shape  of  the  open- 
ing in  the  round  pipe  is  exactly  similar  to  that  de- 
scribed in  the  several  preceding  problems. 


PROBLEM    59. 


A  T-Joint  Between  Pipes  of  Different  Diameters,  the  Axis  of  the  Smaller  Pipe  Passing:  to  One  Side  of  That  of  the  Larger. 


The  principle  here  involved  and  the  method  of 
procedure  are  exactly  the  same  as  in  Problem  5<>,  but 
the  whole  of  the  profiles  must  be  used  instead  of  the 
halves,  because  the  two  axes  or  center  lines  of  the 
pipes  do  not  intersect. 

In  Fig.  347,  let  A  B  C  be  the  size  of  the  small 
pipe  and  F1  H1  M1  be  the  size  of  the  large  pipe,  between 
which  a  right-angled  joint  is  to  be  made,  the  smaller 

i 


pipe  being  set  to  one  side  of  the  axis  of  the  large  pipe, 
as  indicated  in  the  end  view.  Draw  an  elevation,  as 
shown  by  D  F  I  L  M  K  G  E.  Place  a  profile  of  the 
small  pipe  above  each,  as  shown  by  ABC  and  A'  B' 
C1,  both  of  which  divide  into  the  same  number  of  equal 
parts,  commencing  at  the  same  point  in  each.  Place 
the  T-square  parallel  to  the  small  pipe,  and,  bringing 
it  successively  against  the  points  in  the  profile  A1  B' 


Pattern   Problems. 


157 


C',  drop  lines  cutting  tin-  profile  of  the  large  pipe,  as 
shown,  from  F1  to  II';  and  in  like  manner  drop  lines 
from  the  points  in  the  profile  A  B  C,  continuing  them 
through  the  elevation  of  the  larger  pipe  indefinitely. 
For  the  pattern  of  the  small  pipe  set  off  a  stretchout 


Ai 


-  -12 


13  R 


Fig.  .Itf.—A  t-Joint  between  Pipes  of  Different  Diameters,   the  Axis 
of  the  Smaller  ripe  Passing  to  One  Side  of  that  of  the  Larger. 


line,  V  W,  at  right  angles  to  and  opposite  the  end  of 
the  pipe,  and  draw  the  measuring  lines,  as  shown. 
These  measuring  lines  are  to  be  numbered  to  corre- 
spond to  the  spaces  in  the  profile,  but  the  place  of 
beginning  determines  the  position  of  the  seam  in  tin- 


pipe.  In  the  illustration  giv.  .  the  seam  lias  been 
located  at  the  shortest  part  of  the  pipe,  or,  in  other 
words,  at  the  line  corresponding  to  the  point  10  in  the 
section.  Therefore  commence  numbering  the  stretchout 
lines  with  10.  Place  the  T-square  at  right  angles  to  the 
small  pipe,  and,  bringing  the  blade  suc- 
cessively against  the  points  in  the  pro- 
file of  the  large  pipe  from  F1  toH',  cut 
the  corresponding  measuring  lines,  as 
shown.  A  line  traced  through  the  points 
thus  obtained,  as  shown  by  X  Y  Z,  will 
form  the  end  of  the  required  pattern. 

For  the  pattern  of  the  large  pipe, 
lay  off  a  stretchout  from  the    profile 
shown  in  the  end  view,  beginning  the 
same  at  whatever  point  it  is  desired  to 
locate  the  seam,  which  in  the  present 
instance  will  be  assumed    on    a    line 
corresponding  to   point   13  in  the  pro- 
•  file.      After    laying    off    the  stretch- 
out   opposite    one    end    of    the    pipe, 
as  shown  on  0  R,  draw  a  corresponding  line  opposite 
the  other,  as  shown  by  N  P,  and  connect  N  0  and  P 
R,  thus  completing  the  outline  of  the  pattern,  through 
which  an  opening  must  be  cut  to  miter  with  the  end 
of  the  smaller  pipe.     In  spacing  the  profile  of  the  large 
pipe,  the  spaces  in  that  portion  against  which  the  small 
pipe  fits  are  made  to  correspond  to  the  points  obtained 
by  dropping  lines  from  the  profile  of  the  small  pipe 
upon  it,  as  shown  by  1  to  7  inclusive.      This  is  done 
in  order  to  furnish  points  in  the  stretchout  correspond- 
ing to  the  lines  dropped  from  the  profile  A  B  C,  as 
shown.     No  other  measuring  lines  than  those  which 
represent  the  portion  of  the  pipe  which  the  small  pipe 
fits  against  are  required  in  the  stretchout.      Accord- 
ingly the  lines  1  to  7  inclusive  are  drawn  from  0  R, 
as  shown,  and  are  cut  by  corresponding  lines  dropped 
from  ABC.      A  line  traced  through  the  several  points 
of  intersection  gives  the  shape  S  T  U,  which  is  the 
opening  in  the  large  pipe.      If  it  be  necessary  for  any 
purpose  to  show  a  correct  elevation  of  the  junction  be- 


tween two  pipes, 


the  miter  line  F  H  G  is  obtained  by 
intersecting  the  lines  dropped  from  ABC  with  cor- 
responding lines  carried  across  from  the  .same  points 
obtained  on  the  profile  F1  IF,  by  dropping  from  A'  B' 
C',  explained  in  Problems  ;">(>,  ;">7  and  5.S,  and  all  as 
shown  by  the  dotted  lines. 

As  remarked  in  Problem  .">•'.,  this  line  is  not  abso- 
lutely necessary,  but  is  of  great  advantage  in  illustrat- 
ing the  nature  and  principles  of  the  work  to  be  done. 


158 


TIte  Xcw  Metal    \\'u/-L-,'f   I  \iltrn,    !',<>,, I. 


PROBLEM    60. 

A  Joint  at  Other  Than  Right  Angles  Between  Two  Pipes  of  Different  Diameters,  the  Axis  of  the  Smaller  Pipe 

Being  Placed  to  One  Side  of  That  of  the  Larger  One. 


In  Fig.  348,  let  C'  B1  A'  be  the  size  of  the  smaller 
pipe,  and  D'  E1  I1  the  size  of  the  larger  pipe,  between 
which  a  joint  is  required  at  an  angle  represented  In  \V 
F  K,  the  smaller  pipe -to  be  placed  to 
the  side  of  the  larger.  Draw  an  eleva- 
tion of  the  pipes  joined,  as  shown  by 
V  D  G  H  I  K  F  W.  As  in  the  preced- 
ing problems,  the  miter  line  or  line  giving 
a  correct  elevation  of  the  junction  of 
the  pipes  must  be  developed  before  the 
actual  work  of  laying  out  the  miter  pat- 
terns can  be  begun,  therefore  place  a 
profile  or  section  of  the  arm  in  line  with 
it,  as  shown  by  C1  B'  A1,  and  opposite 
and  in  line  with  the  end  of  the  main 
pipe  draw  a  section  of  it,  as  shown  by 
D'  E1  I1.  Directly  above  this  section 
draw  a  second  profile  of  the  small  pipe, 
as  shown  by  C  B  A,  placing  the  center 
of  it  in  the  required  position  relative 
to  the  center  of  the  profile  of  the  large 
pipe.  Divide  the  two  profiles  of  the 
small  pipe  into  the  same  number  of  equal  spaces, 
commencing  at  the  same  point  in  each.  From  the 
divisions  in  C'  B'  A'  drop  lines  parallel  to  the  lines  of 
the  arm  indefinitely.  From  the  divisions  in  C  B  A 
drop  lines  until  they  cut  the  profile  of  the  large  pipe, 
as  shown  by  the  points  in  the  arc  D'  E1.  From  these 
points  carry  lines  horizontally  to  the  left,  producing 
them  until  they  intersect  the  corresponding  lines  from 
C'  B1  A1.  A  line  traced  through  these  points  of  in- 
tersection, as  shown  by  D  E  F,  will  be  the  miter  line 
between  the  two  pipes.  For  the  pattern  of  the  arm 
proceed  as  follows :  Lay  off  a  stretchout  at  right 
angles  to  and  opposite  the  end  of  the  arm,  as  shown 
by  R  P,  and  through  the  points  in  it  draw  the  usual 
measuring  lines.  Place  the  T-square  at  right  angles 
to  the  arm,  and,  bringing  it  successively  against  the 
points  already  in  the  miter  line,  cut  the  correspond- 
ing measuring  lines.  A  line  traced  through  these 
points,  as  shown  by  UTS,  will  form  the  required 
pattern. 

For  the  pattern  of  the  main  pipe  draw  a  stretch- 
out line  Opposite  one  end  of  it,  as  shown  by  M  0, 
numbering  the  divisions  in  it  with  reference  to  locat- 
ing the  seam,  which  can  be  placed  at  any  point 


desired.  The  spaces  of  the  profile  between  D'  and  K' 
should  be  transferred  to  the  stretchout  point  by  point 
as  they  occur,  as  by  so  doing  measuring  lines  will  be 


Fig.  348  —A  Jnint  at  other  than  Right  Angles  between  Two  Pipes  of 
Different  Diameters,  the  Axin  of  the  Smaller  Pipe  being  Placed  lt> 
One  Side  of  that  of  the  Lartjtr  One. 

obtained  which  will   correspond  to  the  points  already 
in    the  miter   line.      Draw  a  line  corresponding   to  the 


Pattern     I'roliterns. 


159 


stretchout  line  opposite  the  other  end  of  the  pipe,  as 
shown  by  L  N,  and  connect  L  M  and  N  0.  Through 
the  points  in  the  stretchout  line  corresponding  to  the 
points  between  D1  and  E'  of  the  profile  draw  measuring 
lines,  as  shown  by  10,  11,  12,  13,  14,  15  and  1.  Place 
the  T-square  at  right  angles  to  the  main  pipe,  and,  bring- 


ing the  blade  against  the  points  in  1)  K  K  successively, 
cut  the  measuring  lines  of  corresponding  number,  all 
as  shown  by  the  dotted  lines.  A  line  traced  through 
these  points  of  intersection,  as  shown  near  the  middle 
of  M  L  N  0,  will  give  the  shape  of  the  opening  to  be 
cut  in  the  pattern  of  the  main  pipe. 


PROBLEM  61. 


The  Patterns  for  a  Pipe  Intersecting  a  Four-Piece  Elbow  Through  One  of  the  Miters. 


In  Fig.  34!),  lot  A  B  C  1)  E  E1  U1  C'  13'  A'  repre- 
sent the   four-pieced  elbow   in  elevation,  F  G  II  J  its 


Fig.  S49. — Pattern  for  the  Pipe  Intersecting  an   Elbow   Tliroiigh  Our. 
of  the  Miters. 


prolile,  and  K  L  M  X  the  elevation  of   the  pipe  which 
intersects  the  elbow  through  a  miter  joint.     In  line  with 


the  pipe  draw  the  profile  of  sftme,  as  indicated  by  0  P 
R  S.  Extend  F  H  of  the  profile  of  elbow,  upon  which 
as  a  center  line  draw  another  profile  of  the  small  pipe, 
as  shown  by  0'  P'  R'  S1.  Divide  both  profiles  of  the 
small  pipe  into  the  same  number  of  parts,  commencing 
at  the  same  points  in  each,  as  S  and  S1.  Now  parallel 
to  F  P1  of  the  profile  draw  lines  from  the  points  in  O' 
P1  R'  S1  intersecting  the  profile  F  G  II  J.  as  shown. 

A  profile  should  properly  be  drawn  in  its  correct 
relation  to  the  part  of  which  it  is  the  section.  As  the 
part  C  D  D1  C'  is  about  to  be  considered  first,  the  pro- 
file should  be  pjaced  with  its  center  line  F  H  at  right 
angles  to  C  D ;  but  as  in  a  regular  elbow  of  any  num- 
ber of  pieces  the  miter  lines  all  bear  the  same  angle 
with  the  sides  of  the  adjacent  pieces,  the  profile  may 
for  convenience  be  placed  in  proper  relation  to  one  of 
the  end  pieces,  after  which  lines  may  be  carried  from 
it  parallel  to  the  side  it  represents  to  the  miter  line, 
thence  from  one  miter  line  to  another,  always  keeping 
parallel  to  the  side,  continuing  this  throughout  the 
entire  elbow  if  necessary.  Therefore  parallel  to  D  E 
of  the  elevation  draw  lines  from  the  points  in  G  H  of 
the  profile,  cutting  the  miter  line  D1  D,  and  continue 
these  lines  parallel  to  D  C  and  C  B.  From  the  points 
in  the  profile  0  P  R  S  draw  lines  parallel  with  L  K 
intersecting  lines  of  corresponding  numbers  drawn  from 
G  II.  A  line  traced  through  these  intersections  will 
give  the  miter  line  K  Z  N.  From  the  point  Z  in 
the  miter  line  carry  a  line  back  to  the  profile  of  the 
pipe,  as  indicated  by  Z  a.  This  gives,  upon  the  pro- 
file of  the  pipe,  the  point  at  which  the  rniter  line  K  N 
crosses  the  rniter  line  of  the  elbow  C  C1,  so  that  it  can 
be  located  upon  the  stretchout  line,  where  it  is 
marked  «'. 

For  the  pattern  of  the  pipe  K  L  M  N  proceed  as 
follows:  At  right,  angles  to  K  L  draw  the  line 
M'  M",  upon  which  lav  oil'  the  stretchout  of  ()  I' 
1!  S.  as  shown  bv  the  smiill  figures,  through  the 
points  in  which,  and  at  right  angles  to  it,  draw  the 


100 


The    \'t'tn    Mini     \\~vrkcr   Pattern    liuuk. 


usual  measuring  lines,  which  intersect  with  lines  of 
corresponding  numbers  drawn  at  right  angles  to  the 
line  of  the  pipe  L  K  from  the  intersections  on  the 
miter  line  K  Z  N.  A  line  traced  through  the  inter- 
sections thus  obtained,  as  shown  by  M1  N1  K1  N2  M", 
will  be  the  required  pattern  for  the  intersecting  pipe. 

To  avoid  confusion  of  lines  in  developing  the 
patterns  of  the  intersected  pieces  of  the  elbows  a 
duplicate  of  those  parts,  as  shown  by  B  C  D  D'  C1  B1, 
is  given  in  Fig.  350,  in  which  the  miter  line  K  Z  N  is 
also  shown.  The  profiles  F  G  H  J  and  O'  P'  K'  S'  are 
presented  merely  to  show  the  relationship  of  parts,  as 
the  patterns  are  obtained  from  the  miter  line  K  Z  N, 
in  connection  with  the  stretchout  of  as  much  of  the 
profile  as  is  covered  by  the  intersection.  It  is  not 
necessary  to  include  in  this  operation  the  entire  elbow 
pattern,  therefore  only  such  a  part  of  the  pattern  will 
be  developed  as  is  contained  in  profile  from  V  to  H. 

For  the  pattern  for  that  portion  of  elbow  shown 
in  elevation  by  U  Z  N  or  V  II  of  profile,  proceed  as 
follows :  At  right  angles  to  C  D  of  elevation  draw 
the  line  R  S,  upon  which  lay  off  the  stretchout  of  V  H 
of  the  profile,  as  indicated  by  the  small  figures,  through 
which  draw  the  usual  measuring  lines  at  right  angles 
to  it,  which  intersect  with  lines  of  corresponding  num- 
bers drawn  from  the  intersections  on  the  miter  line 
Z  N  at  right  angles  to  C  D.  Trace  a  line  through  the 
intersections  thus  obtained,  as  shown  by  U3  Z3  N1. 
Then  will  U3  Zs  represent  the  pattern  for  that  part  of 
elbow  shown  in  elevation  by  U  Z,  and  Z3  N1  be  the 
pattern  for  the  cut  on  the  miter  line  Z  N.  The  pat- 
tern for  the  other  half  of  opening  shown  by  N1  X  V  S 
is  simply  a  duplicate  of  the  half  just  obtained  re- 
versed. Then  X  N'  Z"  shows  the  shape  to  be  cut  out 
of  what  would  otherwise  be  a  regular  elbow  pattern. 
The  point  a  in  the  profiles  O'  S'  and  V  II  is  so  near 
the  line  drawn  from  the  point  4  that  separate  lines 
are  not  shown,  and  on  this  account  when  obtaining 
the  shape  of  K1  Z'  the  points  4  and  a  are  shown  on  the 
same  line. 

In  order  to  show  that  the  pattern  is  produced  by 
the  regular  method — -that  is,  by  the  intersection  of 
points  from  the  miter  lino  into  lines  of  corresponding 
number  in  the  stretchout — it  should  be  noted  that  the 


profile  and  Stretchout  of  the  piece  alivadv  developed 
is'  properly  designated  by  the  figures  1,  '2,  3,  «,  5,  (i, 
while  that  of  the  piece  next  to  be  considered  is 
properly  designated  by  the  figures  1,  2,  3,  -ia,  5,  6,  the 
point  4  not  occurring  in  the  first  piece  at  all,  while  the 
points  i  and  a  both  fall  upon  the  same  line  in  the  stretch- 
out of  the  second  piece,  all  of  which  is  clearly  shown. 
The  pattern  of  the  cut  on  the  miter  line  K  Z  is 
obtained  in  the  same  manner  as  for  Z  N.  At  right 
angles  to  B  C  draw  the  line  R'  S',  upon  which  place 
the  stretchout  of  V  H  of  the  profile,  as  shown. 


s1 


B1  D 

ELEVATION 


Fig.  S50. — Patterns  for  the  Pieces  of  an  Elbow  Intersected  by  the 

Pip*. 

Through  the  points  in  the  stretchout  and  at  right 
angles  to  same  draw  the  usual  measuring  lines,  which 
intersect  with  lines  of  corresponding  numbers  drawn 
from  the  intersections  on  the  miter  line  K  Z,  at  right 
angles  to  B  C.  Trace  a  line  through  the  intersections 
thus  obtained,  as  shown  by  K1  Z'  U'.  Then  will 
Z'  U'  represent  the  pattern  for  that  part  of  elbow 
shown  in  elevation  by  Z  U,  and  Z1  K'  be  the  pattern 
for  the  cut  on  the  miter  line  Z  K.  The  pattern  for 
the  other  half  of  opening  shown  by  K'  X'  V  S'  can  be 
obtained  by  duplication.  Then  will  X1  K'  Z'  repre- 
sent the  shape  to  be  cut  out  of  the  regular  clb<>\\- 
pattern. 


Pattern  Problems. 


161 


PROBLEM   62. 

The  Pattern  for  a  Gable  Molding:  Mitering  Against  a  Molded  Pilaster. 


Let  N  X  V  E  in  Fig.  351  be  the  elevation  of  a 
<r:ililc  molding  of  which  A  BCD  is  the  profile,  and 
K  (.)  M  L  be  the  elevation  of  a  molded  pilaster  against 


straight  from  N  to  E,  as  would  be  the  case  if  the  side 
of  the  pilaster  were  perfectly  flat  and  projected  further 
than  the  gable  molding.  It  will  therefore  be  neces- 


Fig.  151.— The  Pattern  for  a  Gable  Molding  Mitering  Against  a  Molded  Pilaster. 


which  it  is  required  to  miter.  The  profile  of  the 
pilaster  is  shown  by  J  I  H  in  the  plan,  where  a  profile 
of  the  gable  mold  A1  B1  C1  D1  is  also  shown  and  so 
placed  as  to  show  the  comparative  projection  of  the 
various  points  in  each.  By  an  inspection  of  the  plan 
and  elevation  it  will  be  seen  that  the  miter  line  or  joint 
between  the  molding  and  the  pilaster  will  not  be 


sary  to  first  obtain  a  correct  elevation  of  the  miter, 
after  .which  the  pattern  can  be  obtained  in  the  usual 
simple  manner. 

To  do  this  divide  the  profiles  in  the  plan  and 
elevation  into  the  same  number  of  equal  parts,  com- 
mencing at  the  same  points  in  each,  as  shown  by  the 
corresponding  figures,  From  the  divisions  in  the  pro- 


162 


The   Xeir  Metal    Worker    /'nttrn,    /AW,-. 


file  in  plan  carry  lines  to  the  left,  parallel  to  H  A', 
until  they  cut  the  side  of  the  pilaster  II  I  J,  as  shown. 
From  these  intersections  drop  lines  at  right  angles  to 
H  A1  indefinitely,  as  shown.  From  the  divisions  in 
the  profile  in  elevation  draw  lines  parallel  to  N  X 
until  they  intersect  corresponding  lines  drawn  from 
II  I  in  plan.  A  line  traced  through  these  intersections, 
as  shown  by  N  F  E,  will  be  the  required  miter  line,  or 
intersection  of  the  gable  molding  with  the  upright 
pilaster  at  the  angle  0  N  X. 

Foi  the  pattern  of  the  gable  molding  proceed  as 
follows :  At  right  angles  to  the  lines  of  the  gable 
molding  draw  the  stretchout  line  A"  D",  upon  which 
place  the  stretchout  of  the  profile  of  the  gable  mold- 
ing, through  which  draw  the  usual  measuring  lines, 
which  intersect  with  lines  of  corresponding  number 
drawn  from  the  points  in  the  miter  line  N  F  E  at  right 


angles  to  the  lines  of  the  gable  molding.  A  line  traced 
through  these  intersections,  as  shown  1>\-  N  G  E2,  will 
be  the  required  pattern. 

Although  the  roof  strip  A  B  of  the  gable  mold- 
ing is  perfectly  straight,  points  will  have  to  be  intro- 
duced between  A  and  B  for  the  purpose  of  ascertaining 
the  shape  of  the  cut  from  N  to  F,  its  intersection  with 
the  side  of  pilaster.  The  simplest  method  of  obtain- 
ing these  points  is  to  derive  them  from  the  points  be- 
tween B'  and  C1,  as  shown  by  0'  to  5'  in  the  plan. 
They  can  then  be  transferred  to  their  proper  place  in  the 
stretchout,  as  shown,  between  A3  and  B".  By  so  doinir 
points  of  like  number  fall  in  the  same  place  on  the  pro- 
lile  H  I,  and  the  vertical  lines  dropped  therefrom  ran  !><• 
intersected  with  F  X  for  the  pattern  of  the  roof  strip 
and  with  the  other  lines  from  F  to  E  for  the  pattern  of 
the  face  of  the  mold,  all  of  which  is  clearly  shown. 


PROBLEM    63. 

The  Patterns  for  an  Anvil. 


It  frequently  occurs  that  sheet  metal  reproductions 
of  various  emblems  or  tools  are  desired  for  use  as  orna- 
ments or  signs.  In  the  following  problem  is  shown 
how  the  various  pieces  necessary  to  form  an  anvil  may 
be  obtained.  The  description,  of  course,  only  applies 
to  the  several  sides,  as  a  representation  of  the  horn 
can  only  be  obtained  by  hammering  or  otherwise  stretch- 
ing the  metal. 

In  Fig.  352  is  shown  a  side  and  end  elevation  and 
two  plans  of  the  anvil,  exclusive  of  the  horn,  the  plans 
being  duplicates  and  so  placed  as  to  correspond  re- 
spectively with  the  side  and  the  end  views.  Before 
the  pattern  of  the  side  piece  J  N  B  G  can  be  developed 
the  line  0  P  Q  R  S,  which  is  the  result  of  the  miter- 
ing  together  of  the  two  forms  shown  by  U  V  W  of  the 
plan  and  Z  X  T  of  the  end  view,  must  be  obtained. 
Therefore  divide  the  curved  portion  Z  X  T  of  the 
profile  of  the  side  into  any  convenient  number  of 
equal  spaces,  as  shown  by  the  small  figures,  and  from 
the  points  thus  obtained  drop  lines  vertically  cutting 
W  V  U1,  the  profile  of  the  gore  piece.  Transfer  the 
points  thus  obtained  on  W  V  U'  to  W  V  U  of  the 
other  plan,  and  from  these  points  erect  lines  vertically 
through  the  elevation  of  the  side,  and  finally  intersect 
them  with  lines  of  corresponding  number  drawn  from 
the  points  originally  assumed  in  Z  T.  Then  a  line 
traced  through  the  points  of  intersection,  as  shown  by  ' 
0  P  Q  R  S,  will  be  the  required  miter  line. 


To  obtain  the  pattern  of  the  side,  first  lay  off  a 
stretchout  of  the  profile  Z  T,  as  shown,  upon  v  q,  through 
the  points  in  which  draw  the  usual  measuring  lines. 


K 

J 

N 
M 

\ 

^ 

N1 
M' 

I~\K' 

B 
E 

c 

B' 
E' 

ft? 

1  l\< 

2  
3  

2  

l    1 

I     |     ELEV/ 

TION       1     | 

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TION     |  |      | 

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1  1 

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D 
Fig.  S5S.— P.itterns  for  the  End  Pieces  of  an  Anvil. 

With  the  T-square  placed  parallel  with  J  G  drop  lines 
from  the  points  in  the  profile  Z  T  cutting  the  outlines. 


Pattern   Problems. 


163 


of  the  side  from  J  to  a  and  (i  to  //,  and  also  cutting  the 
miter  lines  of  the  gore  piece  O  Q  S  (which  last  opera- 
tion has  really  been  done  in  the  raking  operation  above 
described).  Placing  the  T-square  parallel  with  v  q, 
bring  it  successively  against  the  points-  in  the  several 
miter  lines  of  the  side  elevation  and  cut  corresponding 
measuring  lines;  then  lines  traced  through  the  points 


the  raking  operation,  lay  off  a  stretchout  of  the  same 
upon  any  line  running  at  right  angles  to  the  form  of 
this  piece,  as  shown  upon  U"  W.  As  the  points  have 
already  been  dropped  from  the  profile  to  the  miter  lines 
in  the  operation  of  obtaining  them,  it  only  remains  to 
place  the  I  -square  parallel  to  IP  W*  and  bring  it  suc- 
cessively against  the  points  in  0  Q  and  Q  S.  cutting 


|| I  (ELEVATION  jSIDE  | 


Fig.  352. — Patterns  for  the  Side,   Gore  Piece  dud  Bottom  of  an  Anvil. 


of  intersection,  as  shown,  from  y  to  e,  o  to  X,  X  to  s  and 
g  toe?,  will  give  the  pattern  for  the  lower  portion  of  the 
side.  As  that  part  of  the  side  from  Y  to  Z  of  the  pro- 
file is  straight  and  vertical,  that  portion  of  the  pattern 
shown  on  the  stretchout  line  from  X  to  </  can  be  made 
an  exact  duplicate  of  that  part  of  the  elevation  shown 
by  a  M  N  B  E  b,  all  as  shown. 

For  the  pattern  of  the  gore 'piece,  U  V  W  is  the 
profile  and  O  Q  and  Q  S  are  the  miter  lines.  By  means 
of  the  points  previously  obtained  upon  the  profile  in 


corresponding  measuring  lines;  then  lines  traced 
through  the  points  of  intersection,  as  shown  by  U1 
Q1  and  Q1  W,  will  give  the  pattern  for  the  gore 
piece. 

For  the  end  pieces  of  the  anvil,  N  M  a  K  J  and 
B  E  b  H  G  of  the  side  elevation  become  the  profiles, 
and  Z  T  and  Z1  T1  are  the  miter  lines.  Therefore,  to 
obtain  the  pattern  of  either  of  these  pieces,  inde- 
pendently of  the  preceding  operations,  space  the  curved 
portion  of  its  profile  into  any  convenient  number  of 


164 


The  New  Metal    Worker  Pattern    Booh. 


spaces,  and  lay  off  a  stretchout  of  the  same  upon  any 
line  at  right  angles  to  T  T'.  Carry  lines  from  the 
profiles  parallel  to  N  B,  cutting  the  miter  lines,  thence 
at  right  angles  to  T  T',  cutting  corresponding  measur- 
ing lines.  To  avoid  confusion  of  lines  the  operation 
of  obtaining  the  patterns  of  the  end  pieces  has  been 
shown  separately  in  Fig.  353,  in  which  J  N  N'  J1  is 
an  elevation  of  the  front  end  and  G  B  B1  G1  that  of 
the  back  end.  The  points  made  use  of,  however, 
upon  their  profiles,  in  Fig.  352,  are  such  as  were  ob- 


tained there  in  cutting  the  pattern  of  the  side;  therefore 
their  stretchouts  must  be  transferred  point  bv  point  to 
the  stretchout  lines;  E  D  of  Fig.  353  being  the  stretch- 
out of  N  a  J  in  Fig.  352  and  B  A  of  Fig.  353  being 
that  of  B  b  G.  In  consequence  of  the  above  the  points 
upon  the  miter  line  Z  T  are  such  as  were  originally 
obtained  there  by  spacing,  and  have  been  transferred 
to  the  lines  N  J,  N1  J1,  B  G  and  B1  G'.  The  remainder 
of  the  work  is  shown  sufficiently  clear  to  need  no 
further  explanation. 


PROBLEM   64. 
The  Pattern  for  a  Gable  Cornice  Mitering  Upon  an  Inclined  Roof. 

In  Fig.  354   let   ABCG    represent    one   side   of      several   points   in   A    G    and    0    B,    cut   corresponding 


the  gable  molding  and  N  O  M  its  profile.  II  B  F  E 
represents  the  horizontal  molding,  and  I)  C  the  upper 
line  of  roof.  The  profile  of  this  molding  is  shown  bv 
K  L,  and  the  inclined  roof  by  K  J.  Before  the  pat- 
tern for  gable  can  be  described  it  will  be  necessarv  to 
obtain  an  elevation  of  the  intersection  of  the  gable 
cornice  with  the  inclined  roof  between  C  and  B  to  be 
used  as  the  miter  line. 

The  first  step  to  be  taken  in  obtaining  this  mitej- 
line  is  to  draw  the  profile  of  gable  cornice,  P  Q  R, 
directly  over  and  in  line  with  the  profile  of 
the  horizontal  molding  J  K  L,  as  shown. 
Divide  both  profiles  O  M  and  P  R  into  the 
same  number  of  parts.  From  the  points  in 
0  M  draw  lines  parallel  with  the  rake,  ex- 
tending them  indefinitely  in  the  direction  of 
C  B.  From  the  points  in  P  R  drop  lines 
upon  the  roof  line  J  K,  and  from  the  points 
of  intersection  in  J  K  carry  lines  liorizontallv 
across  to  the  elevation,  intersecting  them  with 
lines  of  corresponding  number  previously 
drawn  from  0  M.  Through  the  points  of 
intersection  trace  a  line,  which  will  be  at  once 
the  correct  elevation  of  the  miter  and  the 
miter  line  from  which  to  obtain  the  pattern. 
If  the  pattern  of  gable  at  A  G  is  required,  in 
connection  with  that  at  the  foot,  extend  the 
lines  from  points  in  M  O  to  the  miter  line  A  G. 

To  obtain  the  pattern  of  A  B  C  G,  pro- 
ceed as  follows :  At  right  angles  to  A  B  of 
gable  lay  out  a  stretchout  of  M  O,  as  shown  by  S  T, 
through  the  points  in  which  draw  the  usual  measuring 
lines.  Place  the  T-square  at  right  angles  to  the  gable 
line  A  B,  and,  bringing,  it  successively  against  the 


measuring  lines,  all  as    indicated    b 


ELEVATION 

Fig.  S54. — Method  of  Obtiiinimj  Miter  Line  mid  I'ullcrii  for  a  Gable  Cornice 
Milering  Upon  an  Inclined  Roof. 


Thus  the  line  U  X  of  pattern  is  of  the  same'  length  as 
A  B  of  elevation,  and  V  \V  of  pattern  the  same  length 
as  (\  C  of  elevation,  etc.  It  is  evident  that  the  various 
lines  in  pattern  arc  of  the  same  length  as.  lines  of  corre- 


W 


Pattern 


1G5 


sponding   number   in   elevation.      Through   the  points 
obtained  in  the  pattern  trace  lines  as  indicated  by  U  V 


and   W  X.     Then   U  V  W  X  is    the    pattern  for  the 
part  of  gable  shown  by  A  B  C  (i  in  elevation. 


PROBLEM  65. 

The  Pattern  for  the  Molding  on  the  Side  of  a  Dormer  Mitering  Against  the  Octagonal 

Side  of  a  Tower  Roof. 


355.— The  Pattern  for  the  Molding  on  the  Side  of  a  Dormer 
Mitering  Against  the  Octngcm.nl  Side  of  a  Tomer  Roof. 


Let  F  J  H  G  in  Fig.  355  represent  a  half  elevation 
of  a  portion  of  the  tower  roof  corresponding  to  A  B  C 
1)  K  of  the  plan ;  also  let  U  O  P  R  S  be  the  side  eleva- 
tion of  a  dormer  cornice  for  which  the  pattern  is  re- 
quired, K  L  M  N  being  half  the  front  elevation  of  the 
dormer,  and  profile  of  the  molding.  The  first  step 
before  the  pattern  can  be  described  is  to  obtain  a  cor- 
rect elevation  of  the  miter  line  or  intersection  of  the 
cornice  with  the  oblique  side  of  the  tower,  as  shown  by 
UTS.  To  obtain  this  miter  line  proceed  as  follows: 
On  A  E  of  the  plan  extended  as  a  center  line,  as  N1  K1, 
draw  a  duplicate  of  the  half  front  elevation  correspond- 
ing to  the  half  E  D  of  the  plan  as  shown  by  K'  L1  M1 
N1.  Now  divide  the  two  profiles  of  the  return  mold- 
ing into  the  same  number  of  parts,  commencing  at  the 
same  points  in  each,  as  shown  by  the  small  fig.ures. 

From  each  of  the  points  in  the  upper  front  eleva- 
tion carry  lines  parallel  with  U  K  cutting  the  side 
F  J  of  the  tower,  and  for  convenience  in  obtaining  the 
miter  line  extending  them  into  the  figure,  as  shown. 
From  the  intersections  obtained  on  the  side  of  the  tower, 
as  shown  by  the  small  figures  in  U  S,  drop  lines  par- 
allel to  the  center  line  F  G  until  they  cut  the  miter 
line  A  D  of  the  plan. 

From  the  points  in  A  D  draw  lines  parallel  with 
C  D  (the  oblique  side),  extending  them  indefinitely  to- 
ward A  C.  Now  from  the  points  in  the  lower  front  ele- 
vation K1  L1  M1  draw  lines  parallel  with  A  K',  producing 
them  until  they  meet  or  intersect  lines  of  correspond- 
ing numbers,  just  described.  A  line  traced  through 
these  intersections,  as  shown  by  U1  TJ"  T1  S1,  will  give 
the  shape  of  the  miter  line  as 'it  will  appear  in  plan,  TJJ 
T1  S1  showing  that  portion  of  the  intersection  which 
occurs  upon  the  oblique  side  of  the  tower  roof. 

From  the  points  of  intersection  in  the  miter  line 
U"  T'  S1  of  the  plan  erect  lines  parallel  to  A  B, 
producing  them  until  they  intersect  lines  of  correspond- 
ing numbers  drawn  from  the  profile  K  L  M  N  through 
the  side  elevation. 

A  line  traced  through  these  intersections,  as  shown 


New  Metal    Worker  Pattern   Jiook. 


in  U  T  S  of  the  elevation,  will  be  the  miter  line  in  eleva- 
tion, formed  by  the  junction  of  the  return  with  the 
oblique  side  of  the  tower  A  D  C  of  the  plan. 

To  obtain  the  pattern  proceed  as  follows :  At 
right  angles  to  U  0  of  the  elevation  draw  the  line  V 
W,  as  shown,  upon  which  lay  off  the  stretchout  of  K 
L  M  of  the  front  elevation,  as  shown  by  the  small 


figures,  through  which  draw  the  usual  measuring  lines, 
which  intersect  witii  lines  of  corresponding  numbers 
drawn  at  right  angles  to  U  0  of  the  elevation  from  the 
points  of  intersections  in  the  miter  line  U  T  S  and  from 
the  points  of  intersections  in  the  profile  OPE.  A 
line  traced  through  these  intersections,  as  shown  by  U" 
T'  S"  K"  Pa  O',  will  be  the  required  pattern. 


PROBLEM  66. 

The  Pattern  for  an  Inclined  Molding;  Mitering;  Upon  a  Wash  Including1  a  Return. 


As  a  feature  of  design,  it  frequently  occurs  that 
a  belt  course  between  stories  is  carried  around  pilasters 
which  occur  between  all  the  windows  of  a  front,  and 


between  the  pilasters  and  partly  upon  the  wash  of  the 
returns  at  the  sides  of  pilasters.  Such  a  condition  of 
affairs  presents  some  interesting  features  and  is  shown 


PLAN 


Fig.  SOG.—The  Pattern  far  an,  Inclined  Molding  Mitering  Upon  a   Wash  Including  a  Return. 


which  rise  from  the  foundations  to  the  main  cornice. 
In  a  certain  instance,  small  gables  or  pediments  were 
introduced  between  the  pilasters  in  such  a  manner  that 
the  miters  at  the  foot  of  the  gables  came  partly  upon 
the  roof  or  wash  of  that  part  of  the  belt  course  lying 


in  Fig.  356,  of  which  A  B  C  is  the  front  elevation. 
D  E  G  shows  the  plan  of  the  belt  course,  upon  which 
the  foot  of  the  gable  mold  is  required  to  miter  in  the 
vicinity  of  F..  The  gable  mold,  of  which  ,1  C  is  the 
elevation  and  H  the  profile,  is  required  to  meet  the 


Patte 


ir,7 


level  cornice  at  the  angle  C  J  M,  its  top  line  starting 
from  the  point  J.  In  this  instance,  as  in  many  others, 
the  first  requisite  is  that  of  obtaining  a  correct  eleva- 
tion of  the  miter  between  the  gable  mold  and  the  three 
washes.  To  facilitate  this  operation  it  will  be  neces- 
sary to  draw  a  side  view  which  will  show  the  compara- 
tive projection  of  the  gable  mold,  the  pilaster  and  the 
belt  cornice  from  the  face  of  the  wall,  as  shown  at  the 
right.  Divide  both  profiles  of  the  gable  mold  into  the 
same  number  of  spaces,  as  shown  by  the  small  figures. 
From  the  points  in  the  profile  H  of  the  elevation  carry 
lines  parallel  to  C  J,  extending  them  across  the  line 
J  L  indefinitely.  From  the  points  in  H1  drop  lines 
cutting  the  profile  of  the  wash  of  the  belt  course  O  P 
so  far  as  they  fall  within  its  projection.  From  the 
points  in  O  P  carry  lines  horizontally  across  the  eleva- 
tion till  they  intersect  lines  of  corresponding  number 
previously  drawn  from  the  profile  H.  Inspection  will 
show  that  only  the  lower  portion  of  the  profile  will 
miter  upon  the  main  wash  and  that,  therefore,  the 
above  operation  can  be  begun  with  advantage  at  point 
1 4  and  continued  until  a  line  traced  through  the  points 
of  intersection  crosses  the  line  J  K,  which  is  really 
the  profile  of  the  wash  of  the  return.  This,  as  will  be 
seen,  occurs  at  S,  which  point  can  be  carried  back  to 
profile  II  (shown  at  a;),  where  it  will  be  subsequently 
needed  in  obtaining  the  stretchout  of  the  gable  mold. 
Above  the  point  x  of  the  profile  all  points  will  fall 
against  the  wash  of  the  return  J  K  until  the  projec- 
tion of  the  mold  carries  them  across  the  forward  miter 
of  the  return  (J'  K'  in  plan),  after  which  they  will  fall 
upon  the  wash  in  front  of  the  pilaster.  This  point  of 
crossing  can  be  found  by  reversing  the  operation  above 


described,  thus:  From  points  upon  J  K  above  S 
car'y  lines  horizontally  across  to  side  view,  intersecting 
thun  with  lines  of  corresponding  number  dropped 
from  profile  H'  until  a  line  traced  through  points  of 
intersection,  shown  by  S'  T',  crosses  the  line  0'  P1,  as 
shown  at  point  T',  which  point  happens  to  coincide 
with  point  4  of  profile.  From  point  -I  of  profile  H' 
lines  are  dropped  upon  O'  P',  from  which  they  are 
carried  horizontally  across  as  in  the  first  part  of 
the  operation  till  they  intersect  with  lines  of  cor- 
responding number  drawn  from  points  in  profile  H, 
as  shown  from  T  to  N.  Then  the  line  J  N  T  S  L 
will  be  a  correct  elevation  of  the  required  intersection 
and  can  be  used  as  the  miter  line  from  which  to  obtain 
the  final  pattern  of  the  gable  mold.  With  this  as  a 
miter  line  and  II  as  a  profile  the  remaining  operation 
is  performed  in  the  .usual  manner.  Upon  any  line,  as 
V  W,  drawn  at  right  angles  to  J  C  lay  off  a  stretchout  of 
the  profile  of  gable  mold.  In  obtaining  this  stretchout 
the  position  of  point  x  must  be  obtained  from  profile 
H,  while  from  profile  H1  is  obtained  the  position  of 
point  Q,  which  shows  the  point  at  which  the  roof 
piece  of  the  gable  mold  passes  beyond  the  side  of 
the  pilaster,  shown  best  at  J'  in  the  plan.  With  the 
"["-square  placed  parallel  to  V  W,  and  brought  success- 
ively against  the  points  in  J  N  T  S  L,  cut  measuring 
lines  of  corresponding  number.  A  line  traced  through 
the  points  of  intersection,  as  shown  by  J1  N'  S'  L',  will 
give  the  required  pattern.  The  plan  view  of  the  in- 
tersection is  shown  at  N2  L'',  with  some  of  the  lines  of 
projection  used  in  obtaining  it,  merely  to  assist  the 
student  in  seeing  the  relation  of  parts,  but  is  not 
necessary  in  the  actual  work  of  obtaining  the  pattern. 


PROBLEM   67. 


The  Pattern  for  a  Level  Molding;  Mitering  Obliquely  Against  Another  Level  Molding  of  Different  Profile. 


In  Fig.  357  is  shown  the  plan  and  a  portion  of 
the  side  view  of  a  bay  window.  In  the  side  view  is 
also  shown  the  section  of  a  lintel  molding,  shown 
indefinitely  by  C  D  E  F  of  the  plan,  which  it  is  re- 
quired to  miter  against  the  oblique  side  of  the  large 
cove  under  the  bay  window  indicated  by  B  C  F  of  the 
plan.  In  Fig.  358  is  shown  an  enlarged  plan  of  the 
particular  portion  in  which  the  miter  occurs,  the  angle 


BCD  being  the  same  as  B  C  D  of  Fig.  357.  In  Fig. 
358,  A  B  F  G  represents  the  base  of  the  window  and  G  E 
D  C  the  lintel  cornice.  The  profiles  are  shown  respect- 
ively at  Y  and  X.  The  lintel  molding  is  continued  in 
the  direction  of  F  G  until  it  intersects  the  base  of  the 
window  between  G  and  C. 

In  order  to  obtain  the  pattern  of  that  part  of  the 
lintel   molding  which  abuts  against  the   base    of    the 


108 


TV    \en-    Metal     \Vorlrr    Patte 


window  indicated  from  G  to  C,  it  is   first   necessary   to      lino  with  the  profile  Y  draw  a  duplicate  of  X,  as  shown 
obtain  the   plan    of   the   intersection  or  shape  of  the      at  Z.     In  placing  the  profile  Z  in  position  it  must  he 


N,    M 


--4 1- J 1 - 

I           I           I           i  I 

—  + r i 1 1- 


I      III 

54-  -4-44-   -4 L J 1 

III  >          I          I 

-4--  +44 L 1 1 1 

i i i n i 

r 

(. i 1 f      ,£._ 

12 1/ 


Fig.  357, — Kan  and  Sectional  View 
of  a  Level  Molding  Mitering 
Obliquely  Against  Another  Livel 
Molding  of  Different  Profile. 


Fig.  S58. — Method  of  Obtaining  the  Pattern  of  the  Lintel  Molding  Shown  in  Fig.  S57. 

miter  line,  as  shown  in  plan  by  G  H  C  F.     To  obtain      remembered  that  as  nights  are  all  to  be  compared  the 
this  miter  line  proceed  as  follows  :     Opposite  to  and  in      vertical  lines  of  each  profile  must  be  placed  parallel 


I 'ill  III' II    I 'fill  ill' nix. 


Hi!) 


;iinl  their  ii]i|xT  ends  turned  in  tin-  same  direction. 
Therefore,  the  back  or  line  1  \'.\  of  the  profile  /  is 
placed  parallel  to  P>  (',  which  represents  a  vertical  line 
with  reference  to  the  profile  V,  and  the  point  12  is 
placed  exactly  opposite  the  point  .1.  according  to  the 
requirements  of  the  side  view.  Fig.  '••~>7.  Divide  the 
profiles  X  and  /  into  the  same  number  of  parts,  as  in- 
dicated l>v  the  small  figures  in  each.  From  the  points 
thus  obtained  in  profile  /  carry  lines  at  right  angles  to 
P>  ('.  cutting  K  J  of  profile  Y,  as  shown.  With  the 
T-square  placed  parallel  with  the  line  B  (J  of  the  plan 
of  the  window  carry  lines  from  the  points  on  the  pro- 
file K  J  in  the  direction  of  (i  and  F;  also  draw  lines 
from  the  points  in  the  profile  X  parallel  to  Gr  K,  cutting 
the  lines  of  corresponding  munlier  drawn  from  the  pro- 
file Y.  A  line  traced  through  points  of  intersection, 
as  shown  by  (r  H  (<  F,  will  give  the  miter  line,  as 
shown  in  the  plan. 

While  the  curved  portions  of  the  profiles  X  and 
/  have  been  divided  into  such  a  number  of  spaces  as 
will  answer  the  purpose,  of  an  ordinary  miter,  it  will 
be  noticed  that  the  plane  surfaces  between  points  2  and 
.">  and  1.1  and  12  intercept  so  much  of  the  curve  of  K 


.1  as  to  produce  a  curve  between  those  points  in  the 
pattern.  Therefore,  for  accuracy  it  is  necessary  to 
subdivide  those  spaces  on  K  J,  as  indicated  by  a  l>  and 
c  d  e  there  shown.  These  points  must  be  dropped  back 
to  the  profile  Z,  and  the  spaces  thus  produced  transferred 
to  the  stretchout  line  L  M,  all  as  indicated.  The  lines 
indicated  by  the  small  letters  in  K  J  have  only  been 
drawn  partway  in  the  engraving  to  avoid  confusion, 
and  the  measuring  Hues  produced  by  these  points  in 
the  stretchout  have  been  shown  dotted  for  the  sake  of 
distinction.  These  points  are  then  intersected  with  the 
surfaces  to  which  they  belong  in  the  miter  line  G  H 
C,  as  shown  between  2  and  3  and  11  and  12. 

For  the  pattern  of  the  lintel  molding  first  draw  a 
line  at  right  angles  to  it,  as  shown  by  L  M,  on  which 
line  lay  off  a  stretchout  of  the  profile  X,  as  indicated 
by  the  small  figures.  Through  the  points  thus  ob- 
tained draw  the  usual  measuring  lines.  With  the 
T-square  placed  parallel  with  the  stretchout  line  L  M 
j  carry  lines  from  the  points  in  G  H  C  F  to  measuring 
lines  of  -corresponding  number,  when  a  line  drawn 
through  the  points  of  intersection,  as  shown  by  N  0  P, 
will  complete  the  pattern. 


PROBLEM  68. 


The  Patterns  for  a  Square  Shaft  of  Curved  Profile  Mitering  Over  the  Peak  of  a  Gable  Coping  Having 

a  Double  Wash. 


Let  A  B  C  in  Fig  359  be  the  front  elevation  and 
D  E  F  G  H  be  the  side  elevation  of  a  coping  to  sur- 
mount a  gable,  the  profile  of  the  top  of  which  is  shown 
at  D  K  II  in  the  side  elevation.  Also  let  M  N  0  P 
be  the  elevation  of  a  square  shaft  or  base,  as  of  a  finial, 
having  a  curved  profile,  as  shown,  which  it  is  required 
to  miter  down  upon  the  top  of  the  coping.  As  the 
matter  of  drawing  the  elevations  of  the  shaft  in  correct 
position  upon  the  washes  of  the  coping  is  attended  with 
some  difficulty,  the  method  of  obtaining  these  will  be 
briefly  described  first :  Through  the  lowermost  point 
on  either  side  of  the  front  elevation,  as  P,  draw  a  line' 
at  the  correct  angle  of  the  pitch  of  the  coping  or  gable, 
as  shown  by  V  W,  and  extend  the  same  to  the  right 
far  enough  to  permit  a  section  of  the  coping  to  be  con- 
structed upon  it.  Upon  this  line  set  off  the  distance 
X  W,  equal  to  P  L  (half  the  width  of  the  shaft  at  its 


base),  and  through  the  point  W  draw  a  line  perpendicu- 
lar to  V  W ;  next  through  the  point  X  draw  a  line, 
making  the  same  angle  with  Y  W  that  D  K,  of  the 
given  profile  of  the  coping,  does  with  the  horizontal 
line  D  L1,  and  extend  this  line  to  the  right  till  it  meets 
the  line  from  W  at  K',  and  to  the  left,  making  D1  K1 
equal  to  D  K.  This  gives  one-half  the  profile  of  the 
wash,  which  is  all  that  is  necessary  in  obtaining  the 
elevation.  Now  from  the  points  in  the  profile  D'  K1 
project  lines  parallel  to  Y  W  till  they  meet  the  center 
line  of  the  front  elevation,  and  duplicate  them  on  the 
other  side  of  the  center  line,  which  will  complete  the 
front  elevation  of  the  gable.  From  the  points  in  the 
profile  D  K  H  erect  vertical  lines  indefinitely,  which 
may  be  intersected  with  lines  projected  horizontally 
from  the  points  on  center  line  I  L  to  complete  the  side 
elevation.  Thus  a  line  from  point  B  intersected  with 


170 


Tlie  New  Metal    Worker   Pattern  Book. 


line  from  K  will  give  the  apex  of  coping,  and  a  line 
from  point  Z  intersected  with  lines  from  D  and  H  will 
give  the  points  E  and  G,  front  and  back  of  the  washes 
at  the  apex. 

As  the  shaft  is  exactly  square,  the  side  elevation 


point  U ;  while  the  crossing  of  the  side  0  P  with  the 
top  line  of  coping  B  C  (marked  4)  is  projected  upon 
the  center  line  of  the  side  view,  thus  giving  the  point 
Y.  This  completes  the  elevations  with  the  exception 
of  the  lines  M  U  P  and  M1  Y  P1.  If  the  profile  of  the 


of  it,  M1  N'  0'  P1,  is  in  all  respects  the  same  as  that  of  |  shaft  0  P  were  a  straight  line,  either  slanting  or  vertical, 


U4-U 

1  i  i  i 

M--M- 


FRONT  ECEVATION 


Fig.  S59.—The  Patterns  for  a  Square  Shaft  of  Curved  Profile  Mitering  Over  the  Peak  of  a  Gable. 


the  front,  and  is  drawn  in  line  with  it,  as  shown  by  the 
horizontal  lines  of  projection  connecting  them.  This 
having  been  completed,  the  point  at  which  its  side 
M1  N'  crosses  the  line  E  F  (marked  4£)  is  projected 
back  to  the  front  elevation,  as  shown,  thus  locating  the 


the  lines  U  P  and  M1  Y  would  be  straight  lines,  because 
they  would  represent  the  intersection  of  two  plain  sur- 
faces ;  but  as  the  profile  is  a  curved  line  it  will  be 
necessary  to  obtain  correct  elevations  of  these  two  lines, 
as  they  are  essential  in  obtaining  the  patterns  to  follow. 


I'n  n,, -H     I'nJdems. 


171 


The  sliaft  being  square,  the  miters  at  its  angles  are 
plain  square  miters  and  are  developed  by  the  ordinary 
method,  as  explained  in  several  problems  in  the  earlier 
part  of  this  chapter.  The  peculiarity  of  this  problem, 
then,  consists  in  obtaining  the  miter  lines  U  P  and  M1  Y 
and  the  part  of  the  pattern  corresponding  to  the  same, 
which  can  all  be  done  at  one  operation,  as  follows : 

Divide  the  profiles  0  P  and  M'  N'  into  the  same 
number  of  equal  parts,  and  place  the  stretchout  of  the 
same  upon  the  center  lines  extended,  as  shown  at  I  J 
and  I1  J,1  through  which  draw  the  usual  measuring  lines 
for  subsequent  use.  From  the  points  in  M1  N'  from 
4£  down  drop  lines  vertically  upon  the  profile  of 
coping  D  K,  as  shown  by  the  dotted  lines,  and  trans- 
fer the  points  thus  obtained  to  the  profile  D'  K',  from 
which  points  draw  lines  parallel  to  V  W,  as  shown  by 
the  dotted  lines.  Intersect  these  with  lines  of  corre- 
sponding number  (2,  3  and  4)  drawn  horizontally  from 
either  profile,  as  shown  at  2',  3'  and  4',  thus  obtaining 
the  miter  line  U  P.  After  the  points  1  to  11  of 
the  profile  have  been  dropped  into  the  measuring  lines 
of  corresponding  number  of  the  stretchout  the  points  2', 
3',  4'  and  4^'  are  also  dropped  into  the  measuring  lines 
of  corresponding  number,  thus  giving  the  cut  U'  T  at 
the  bottom  of  the  pattern,  which  can  be  duplicated  on 
the  other  side  of  the  center  line,  thus  completing  the 
pattern.  Q  R  S  T  U1  of  the  front  of  the  sliaft. 


From  the  points  1  to  4  of  the  profile  0  P  draw 
lines  parallel  to  B  C  cutting  the  profile  D1  K',  as  shown 
by  the  solid  lines,  and  transfer  the  same  to  the  profile 
of  coping  D  K,  and  from  these  points  erect  perpendicu- 
lar lines  (also  shown  solid)  indefinitely,  as  shown,  which 
intersect  with  lines  drawn  horizontally  from  points  of 
corresponding  number  in  either  profile,  as  shown  at  2," 
3"  and  4".  This  will  give  the  correct  miter  line  M1 
Y.  The  miters  at  the  sides  of  piece  M1  N1  O1  P1 
are  of  course  the  same  as  those  of  the  front  piece, 
therefore  after  they  have  been  obtained  the  points  2" 
and  3"  are  dropped  into  measuring  lines  2  and  3  of  the 
stretchout  I1  J1,  which  when  duplicated  on  the  other 
side  of  the  center  line  complete  the  line  Q'  Y1  T1,  which 
is  the  bottom  cut  of  the  side  piece. 

It  has  been  remarked  that  in  obtaining  the  inter- 
sections between  U  and  P  and  M'  and  Y  horizontal 
lines  may  be  drawn  from  points  in  either  profile.  The 
reason  for  this  is  simply  that  the  two  profiles  O  P  and 
N1  M'  are  identical  and  have  been  divided  into  the 
same  number  of  equal  parts.  If  a  case  should  occur  in 
which  the  side  and  face  should  be  dissimilar  it  must  be 
borne  in  mind  that  N'  M'  is  the  profile  of  the  face  piece 
and  its  points  must  be  used  in  obtaining  the  intersec- 
tions between  U  and  P,  while  O  P  is  the  profile  of  the 
side  piece,  and  its  points  must  be  used  in  obtaining  the 
intersections  between  M1  and  Y. 


PROBLEM  69. 


The  Patterns  of  a  Cylinder  Mitering  with  the  Peak  of  a  Gable  Coping;  Having  a  Double  Wash. 


Let  A  B  C  in  Fig.  360  be  the  elevation  of  a  coping 
to  surmount  a  gable,  the  profile  of  which  is  D  E  F  E1 
D1,  which,  as  will  be  seen,  shows  a  double  wash,  E  Y 
and  F  E1.  Let  M  0  P  N  be  the  elevation  of  a  pipe  or 
shaft  which  is  required  to  miter  over  this  double  wash 
at  the  peak  of  the  gable.  Before  any  patterns  can  be 
developed  it  will  be  necessary  to  first  obtain  a  correct 
elevation  of  the  miter  line  or  intersection  of  the  shaft 
with  the  coping.  To  accomplish  this  proceed  as  fol- 
lows :  In  line  with  the  pipe  or  shaft  construct  a  profile 
of  the  same,  as  shown  by  G1  L1  K1  H1,  which  divide 
into  any  convenient  number  of  equal  parts,  and  from 
the  points  thus  obtained  drop  lines  vertically  through 
the  elevation.  Draw  a  corresponding  profile,  as  shown 


by  II  G  L  K,  directly  over  the  profile  of  the  coping, 
all  as  shown,  which  divide  into  the  same  number  of 
equal  parts,  beginning  to  number  at  a  corresponding 
place  in  the  profile,  and  from  the  points  in  it  drop  lines 
on  to  the  profile  of  the  coping,  cutting  the  washes  E  F 
and  F  E1,  and  thence  carry  the  lines  parallel  to  the  lines 
of  the  coping,  producing  them  until  they  intersect  the 
lines  dropped  from  the  profile  G1  L1  K'  II'.  Through 
the  points  of  intersection  thus  obtained  trace  a  line,  as 
shown  from  0  to  P,  then  0  B1  P  will  be  the  miter  line 
in  elevation. 

As  both  halves  of  the  shaft  are  alike  (dividing  on 
the  line  II  L  in  one  profile,  and  on  H'  L'  in  the  other), 
it  is  really  only  necessary  to  use  one-half  of  the  profile, 


172 


Tin-  New 


Worker   /'<///,•/•,/    />W-. 


MS  to  use  both  halves  as  in  the  diagram  requires  the 
additional  work  of  carrying  the  points  from  E  F  E'  to 
the  center  line  B  B'  for  one-half  and  then  down  the 
other  side  of  the  gable  for  the  other  side.  For  the 
pattern  of  the  shaft  proceed  as  follows :  In  line  with 
the  end  M  N  of  the  shaft,  and  at  right  angles  to  it,  lay 
off  a  stretchout  of  the  profile  G'  H1  K1  L1,  as  shown  by 
R  S;  in  ihc  usual  manner,  through  the  points  in  which 
draw  measuring  lines.  Commence  numbering  these 


unifies  to  the  lines  of  one  side  of  the  coping,  as  A  B, 
lay  oil'  a  stretchout  of  the  wash  of  the  coping  K  I1'  K', 
all  as  shown  by  K'J  F'  E3.  In  this  stretchout  line  set 
oil'  points  corresponding  to  the  points  in  E  F  E1,  ob- 
tained by  the  lines  previously  dropped  from  the  profile 
G  H  K  L.  Place  the  J-square  at  right,  angles  to  A  B, 
and,  bringing  it  against  the  points  in  the  miter  line  O 
B1,  cut  lines  of  corresponding  numbers  drawn  through 
the  stretchout  E"  E3,  all  as  indicated  by  the  dotted 


L'9 


Fig   S60. — The  Patterns  of  a  Cylinder  Mitering  with  the  Peck  of  a  Oable  Coping  Having  a  Double  Wnxh. 


measuring  lines  with,  the  figure  corresponding  to  the 
point  at  which  the  seam  is  desired  to  be,  in  this  case  1. 
Place  the  J  square  at  right  angles  to  the  shaft,  and, 
bringing  it  against  the  points  in  the  miter  line  0  B1  P 
cut  the  corresponding  measuring  lines.  Then  a  line 
traced  through  these  points  of  intersection,  as  shown 
by  T  U  V  W  X,  will  be  the  miter  required.  In  case 
it  should  be  desired  to  miter  the  coping  against  the  base 
of  the  Shaft,  the  pattern  for  it  may  be  obtained  from 
the  same  lines  in  the  following  manner:  At  right 


lines.  Then  a  line  traced  through  these  points  of  in- 
tersection, as  shown  by  Z  1  Y,  will  be  the  pattern  of 
the  wash  for  the  side  of  the  gable  A  B  required  to 
miter  against  the  base  of  the  shaft. 

Incase  the  design  should  call  for  a  shaft  octagonal 
in  shape,  the  same  general  rules  would  apply.  Less 
divisions,  however,  will  be  required  in  the  profile,  it. 
only  being  necessary  to  drop  points  from  each  of  the 
angles  of  the  octagon,  as  in  the  ease  <>f  Problem  33, 
previously  given. 


1'iitlrni.     /'/•<>/:/> ,,/.-,;  173 

PROBLEM    70. 

A  Butt  Miter  of  a  Molding;  Inclined  in  Elevation  Against  a  Plain  Surface  Oblique  in  Plan. 

Let  A  B  in  Fig.  3(>1  be  the  profile  <>l'  a  given 
cornice,  and  let  E  D  represent  the  rake  or  incline  of 
the  cornice  as  seen  in  elevation.  Let  G  H  represent 
the  angle  of  the  intersecting  surface  in  plan.  The 
lirst  step  in  developing  the  pattern  will  be  to  obtain 
the  miter  line  in  the  elevation,  as  shown  by  E  F.  For 
this  purpose  draw  the  profile  A  B  in  connection  with 
the  raking  cornice,  which  space  in  the  usual  manner, 
as  indicated  by  the  small  figures.  Draw  a  duplicate 
of  this  profile,  as  shown  by  A1  B',  placing  it  in 
proper  position  with  reference  to  the  lines  of  the  plan. 
Space  the  profile  A1  B1  into  the  same  number  of  parts 
as  A  B,  and  through  the  points  thus  obtained  carry 
lines  parallel  to  the  lines  of  the  cornice,  as  seen  in 
plan,  cutting  the  line  G  II,  as  shown.  In  like  man- 
ner draw  lines  through  the  points  in  A  B,  carrying 
them  parallel  to  the  lines  of  the  raking  cornice  in  the 
direction  of  E  F  indefinitely,  as  shown.  Place  the 
T-square  at  right  angles  to  the  lines  of  the  cornice,  as 
shown  in  plan,  and,  bringing  it  against  the  points  of 
intersection  in  the  line  G  H,  carry  lines  vertically, 
cutting  corresponding  lines  in  the  inclined  cornice 
drawn  from  the  profile  A  B.  Through  the  points  of 
intersection  thus  obtained  trace  a  line,  as  shown  from 
E  to  F.  Then  E  F  will  be  the  miter  line  in  elevation, 
formed  by  an  inclined  cornice  of  the  profile  A  B  meet- 
ing a  surface  in  the  angle  shown  by  G  H  in  the  plan. 

At  right  angles  to  the  raking  cornice  lay  off  a 
stretchout  of  A  B  upon  any  line,  as  K  L,  and  through 
the  points  draw  the  usual  measuring  lines,  all  as  shown. 
Place  the  T-square  at  right  angles  to  the  lines  of  the 
raking  cornice,  and,  bringing  it  against  the  several 
points  in  the  miter  line  E  F,  cut  corresponding  meas- 
uring lines  drawn  through  the  stretchout  K  L.  A  line 
traced  through  these  points  of  intersection,  as  shown 
from  M  to  N,  will  be  the  pattern  required. 


B' 


/•'!/.  ,161.— A  Butt  Miter  of  a  Molding  Inclined  in  Elevation  Against 
a.  Plain  Surface  Oblique  in  Plan. 


PROBLEM   71. 


Patterns  for  the  Moldings  and  Roof  Pieces  in  the  Gables  of  a  Square  Pinnacle. 


Fig.    302   shows 


the  elevation  of  one  of  four  sim- 
in a  square  pinnacle.      The  profile 


ilar  gables  occurring 

of   the   molding   is   shown    at   P.      The   first  step  is  to 

obtain    the    miter    line    or    elevation    of    the    miter 


shown  at  K,  from  which  to  derive  the  pattern.  Draw 
the  pnrtile  P  in  the  molding,  as  shown,  placing  it  so 
that  its  members  will  correspond  with  the  lines  of  the 
molding.  Draw  a  second  profile,  P',  in  the  side  view 


174 


The  New  Metal   Worker  Pattern  Book. 


of  the  gable,  placing  it,  as  shown  in  the  engraving, 

so  that  its  members  will  coincide  with  the  lines  of  the 
side  view.  Space  both  of  these  profiles  into  the  same 
number  of  parts  in  the  usual  manner,  and  through 
the  points  thus  obtained  draw  lines  parallel  to  each  of 
the  moldings  respectively,  as  shown,  until  they  inter- 
sect, and  trace  a  line  through  the  points  of  intersec- 
tion, as  shown  at  K.  Then  K  is  the  line  in  elevation 
upon  which  the  moldings  will  miter.  Draw  the  center 
line  O  M,  which  represents  the  miter  at  the  top  of  the 
gable. 

For  the  pattern  of  the  molding  lay  off  a  stretch- 
out of  the  profile  upon  any  line,  as  G  H,  drawn  at 
right  angles  to  the  line  of  the  gable  in  elevation,  as 
shown  by  the  small  figures.  Through  these  points 
draw  measuring  lines,  as  shown.  Place  the  T-square 
parallel  to  the  stretchout  line,  or,  what  is  the  same,  at 
right  angles  to  the  line  of  the  gable,  and,  bringing  it 
successively  against  the  several  points  in  the  miter  lines 
0  M  and  K,  cut  the  corresponding  measuring  lines, 
as  shown,  and  trace  lines  through  the  intersections. 
This  completes  the  pattern  of  the  molding,  to  which 
the  piece  forming  the  roof  may  be  added  as  follows : 
Make  L  D1  equal  to  E  D  of  the  side  view  of  the 
gable  and  set  it  off  at  right  angles  to  L  B'.  In  like 
manner,  at  right  angles  to  the  same  line,  set  off  A'  B1 


equal  to  A  B-  of    the  side  view,  and   draw   the  lines 
D'  A'  and  A1  F,  as  shown. 


Fitj.  $62. — Patterns  for  the  Moldings  and  Roof  Pieces  in  the  Gables 
of  a  Square  Pinnacle. 


PROBLEM   72. 


Pattern  for  the  Moldings  and  Roof  Pieces  in  the  Gables  of  an  Octagon  Pinnacle. 


Fig.  363  shows  a  partial  elevation  and  a  portion 
o,f  the  plan  of  an  octagon  pinnacle  having  equal  gables 
on  all  sides.  The  first  step  in  developing  the  patterns 
is  to  obtain  a  miter  line  at  the  foot  of  the  gable,  as 
shown  by  L.  To  do  this  proceed  as  follows  :  Draw 
the  profile  K,  as  shown,  placing  it  so  that  it  shall  cor- 
respond in  all  its  parts  with  the  lines  of  the  molding 
in  elevation.  Divide  into  spaces  and  number  in  the 
usual  manner,  and  through  the  points  draw  lines  par- 
allel to  the  lines  of  the  gable  toward  L,  as  shown. 
Draw  a  duplicate  profile  in  the  plan,  K1,  so  placed  as 
to  correspond  with  the  lines  of  the  molding  in  plan. 
Divide  it  into  the  same  number  of  spaces,  and  through 
the  points  in  it  draw  lines  parallel  to  the  lines  of  the 


plan,  cutting  the  line  D  F,  representing  the  plan  of  the 
miter.  From  the  points  in  D  F  thus  obtained  carry 
lines  vertically,  intersecting  corresponding  lines  dniwn 
from  the  profile  in  the  elevation.  A  line  traced  through 
the  several  points  of  intersection,  as  shown  by  L,  will 
be  the  line  of  miter  in  elevation  between  the  moldings 
of  the  adjacent  gables.  The  center  line  0  N  forms  the 
miter  line  for  the  top  of  the  gable. 

For  the  pattern  proceed  as  follows :  Upon  any 
line,  as  E  E,  drawn  at  right  angles  to  the  lines  of  the 
gable,  lay  off  a  stretchout  of  the  profile,  as  shown  by 
the  small  figures.  Through  the  points  of  the  stretch- 
out draw  the  usual  measuring  lines.  Place  the  T- 
square  at  right  angles  to  the  lines  of  the  gable,  and, 


Pattern   Problems. 


175 


bringing  the  blade  successively  against   the   points  in 
the  two  miter  lines   above  described,    cut   the    corre- 


be  added  by  setting  off  A'  B1  at  right  angles  to  A'  C1, 
equal  in  length  to  A  B  of  the  side  view.     In  like  man- 


F    C 

Fig.  S6S.— Pattern  for  the  Moldings  and  Roof  Pieces  in  the  Gables  of  an  Octagon  Pinnacle. 


sponding  measuring  lines,  as  shown.  Lines  traced 
through  the  points  of  intersection  thus  obtained  will 
give  the  pattern  of  the  molding.  The  roof  piece  may 


ner,  upon  the  line  from  M  set  off  D1  C1  equal  to  C  D 
of  the  side  view.  Then  draw  F'  D1  B',  thus  completing 
the  pattern. 


• 


176  Ttie  Xcw  Metal    Worker  Pattern  Bouk. 

PROBLEM   73. 

The  Pattern  for  the  Miter  Between  the  Moldings  of  Adjacent  Gables  Upon  a  Square  Shaft,  Formed  by 

Means  of  a  Ball. 


In  Fig.  364,  let  A  C  be  one  of  the  gables  in  pro- 
file and  B  D  the  other  in  elevation,  the  moldings  form- 
ing a  joint  against  a  ball,  the  center  of  which  is  at  E. 
The  first  operation  necessary  will  be  that  of  obtaining 
the  miter  line,  or,  in  other  words,  the  appearance  in 
elevation  of  the  intersection  of  the  molding  with  the 
ball.  Place  the  profile  of  the  mold  in  each  gable,  as 
shown  at  F  and  H.  Divide  each  of  these  profiles  into 
the  same  number  of  equal  parts,  as  indicated  by  the 
small  figures.  From  the  points  thus  obtained  in  F 
drop  lines  vertically,  meeting  the  profile  of  the  ball, 
as  shown  from  C  to  J.  From  the  center  E  of  the  ball 
erect  a  vertical  line,  as  shown  by  E  J.  From  the 
points  in  C  J  already  obtained  carry  lines  horizontally, 
cutting  E  J,  as  shown,  and  thence  continue  them,  by 
arcs  struck  from  E  as  center,  until  they  meet  lines  of 
corresponding  number  dropped  from  points  in  the  pro- 
file H  parallel  to  the  gable  in  elevation.  Through  the 
intersections  thus  obtained  trace  a  line,  as  indicated  by 
D  G  M.  Then  D  G  M  will  be  the  miter  line  in  eleva- 
tion. To  develop  the  pattern  for  the  molding,  first  lay 
off  at  right  angles  to  the  gable  a  stretchout  of  the  pro- 
file, as  shown  by  P  R,  through  the  points  in  which 
draw  the  usual  measuring  lines.  Place  the  T-square 
parallel  to  the  stretchout  line,  or,  what  is  the  same,  at 
right  angles  to  the  lines  of  the  gable,  and,  bringing  it 
successively  against  the  points  in  the  miter  line  D  M, 
cut  the  corresponding  measuring  lines.  A  line  traced 
through  the  points  of  intersection  from  2  to  7  (that  is, 
from  U  to  V)  will  give  the  pattern  for  the  curved  por- 
tion of  the  profile. 

As  any  section  of  a  sphere  is  a  perfect  circle  whose 
length  of  radius  depends  upon  the  proximity  of  the 
cutting  plane  to  the  center  of  the  sphere,  the  curves  S 
to  U  and  V  to  T  of  the  pattern,  representing  the  plain 
surfaces  1  2  and  7  8  of  the  profile,  must  be  arcs  of 
circles,  whose  lengths  of  radius  can  be  determined  from 
the  elevation.  As  the  pattern  for  the  plain  surface  1  2 
is  simply  a  duplicate  of  the  cut  from  D  to  G  of  the 


elevation,  set  the  dividers  to  the  radius  E  G  of  the  ele- 
vation, and  from  S  and  U  respectively  as  centers  strike 
arcs,  which  will  be  found  to  intersect  at  N.  Then  N 
is  tin;  center  by  which  to  describe  the  arc  S  U.  To 
find  the  radius  for  the  curve  from  V  to  T  continue  the 
line  8  M  through  the  sphere,  cutting  its  opposite  sides 
at  M  and  L ;  then  M  L  will  be  the  diameter  of  the 


Fig.  S64. — The  Pattern  for  the  Miter  Kftween  the  Moldings  of  Adjacent 
Gables  Upon  a  Square  Shaft,  Formed  bij  Means  of  a  Ball. 

circle  of  which  the  arc  7  8  or  V  T  is  a  part.  There- 
fore with  K  M  (one-half  of  M  L)  as  a  radius,  and  V  and 
T  respectively  as  centers,  strike  arcs,  which  will  inter- 
sect in  the  point  0.  From  0,  with  the  same  radius, 
describe  the  arc  V  T.  Then  S  U  V  T  will  be  the  pat- 
tern, of  the  molding  to  miter  against  the  ball. 


Note. — The  remaining  problems  in  this  section  of  the  chapter  involve  the  necessity  of  "  raking  "  or  de- 
veloping a  new  profile  from  the  given  or  normal  profile  before  the  pattern  for  the  required  part  can  be  ob- 
tained. One  of  the  principal  characteristics  of  this  work  is  that,  as  the  normal  profiles  are  usually  spaced 
into  equal  parts  for  convenience  in  beginning  the  work,  the  resulting  or  raked  profiles  must  by  force  of  cir- 
cumstances be  made  np  of  a  number  of  unequal  spaces  :  in  consequence  of  which  their  stretchouts  must  be 
transferred  to  given  straight  lines,  space  by  space,  as  they  occur  upon  the  new  profiles. 


177 


PROBLEM   74. 

The  Pattern  of  a  Flaring:  Article  of  which  the  Base  is  an  Oblong:  and  the  Top  Square. 


Let  A  B  D  E  of  Fig.  365  be  the  elevation  of  the 
article,  and  F  N  0  I  the  plan  at  the  base,  K  M  P  L 
being  the  plan  at  the  top.  If  the  sides  are  to  be  de- 
veloped in  connection  with  the  top  (supposing  the 
bottom  to  be  open)  proceed  for  the  pattern  as  follows : 
Draw  K'  M'  1"  L',  Fig.  366,  equal  in  all  respects  to 
K  M  P  L  of  the  plan.  Through  the  center  of  it,  and 
at  right  angles  to  each  other,  draw  lines  V  U  and  S  T 
indefinitely.  While  the  elevation  in  Fig.  365  shows 
the  slant  hight  of  the  ends  it  does  not  give  the  slant 


and  L'  P',  making  them  in  length  equal  to  F  N  and 
I  O  of  the  plan,  letting  the  points  V  and  U  come  mid- 
way of  their  lengths  respectively.  Draw  K'  F',  M'  N1, 
L'  I'  and  P1  0',  thus  completing  the  pattern  for  the 
sides.  Upon  S  T  set  off  Z  T  from  M1  P1,  and  Y  S 
from  K1  L',  in  length  equal  to  A  B  or  D  E  of  the  ele- 
vation, and  through  the  points  S  and  T  draw  F'  P  and 
N'  0'  parallel  to  K1  L1  and  M1  P1,  and  in  length  equal 
to  F  I  and  N  0  of  the  plan,  letting  the  points  S  and  T- 
fall  midway  of  their  lengths  respectively.  Draw  FJ  K1 


Fig.  365.— Elevation  and  Plan. 


Fig.  366.— Pattern. 
The  Pattern  of  a  Flaring  Article  uf  Which  the  Base  is  Oblonij  and  the  Top  Square. 


hight  or  profile  of  the  sides,  therefore  through  the  ele- 
vation, and  perpendicular  to  the  base  and  top,  draw 
the  line  C  G,  which  will  measure  the  straight  hight  of 
the  article.  From  G  set  off  G  II,  in  length  equal,  to 
M  R  of  the  plan.  Draw  H  C.  Then  H  C  will  be  the 
profile  of  the  article  through  the  side,  and  therefore 
the  width  of  the  pattern  of  that  portion.  Upon  V  U 
of  the  pattern,  from  K'  M1,  set  off  "W  V,  and  from  L1 
I"  set  off  X  U,  in  length  equal  to  H  C  of  the  eleva- 
tion. Through  U  and  V  draw  lines  parallel  to  K'  M' 


P  L1,  N2  M1  and  O!  P1,  which  will  complete  the  pattern 
of  the  ends. 

If  it  is  required  to  produce  the  pattern  with  the 
sides  joined  to  the  bottom,  supposing  the  top  to  be 
open,  lay  out  first  a  duplicate  of  F  N  I  0,  through  the 
center  of  which  draw  the  stretchout  lines  V  U  and  S 
T  as  before,  and  proceed  in  the  same  general  manner  as 
described  above  to  obtain  the  sides,  placing  then- 
wider  ends  against  corresponding  sides  of  the  base  or 
bottom. 


178 


The  New  Metal   Worker  Pattern  Book. 


PROBLEM  75. 

The  Envelope  of  the  Frustum  of  a  Pyramid  which  is  Diamond  Shape  in  Plan. 


In  Fig.  367,  let  A  B  D  E  be  the  elevation  and 
K  G  I  0  the  plan  of  the  pyramid  at  the  base.  Project 
the  points  B  and  D  into  the  plan,  as  shown,  locating 
the  points  M  and  P,  and  draw  the  sides  of  the  plan  at 
top,  each  parallel  to  the  corresponding  line  of  the 
plan  at  the  base.  By  projection  from  G  or  O  of  the 
plan  draw  C  F  of  the  elevation,  representing  0  R  of 
the  plan  and  also  the  straight  hight  of  the  frustum. 

Before  the  slant  hight  or  stretchout  of  a  side  can 
be  obtained  it  will  be  necessary  to  construct  a  section 
on  any  line  crossing  the  plan  of  the  side  at  right  angles 
as  S  T.  Therefore  extend  the  top  and  bottom  lines 
of  the  elevation,  as  shown  dotted  at  the  right,  cutting 
the  vertical  line  S1  S5,  thus  making  'S1  S"  equal  to  the 
straight  hight  of  the  frustum.  Upon  the  base  line 
extended  set  of  from  S'  the  distance  S1  T1,  equal  to  S  T 
of  the  plan,  and  draw  S3  T1.  Then  will  S"  T1  be  the 
true  profile  or  slant  hight  of  the  frustum. 

At  right  angles  to  M  R  of  the  plan  draw  S  W, 
making  its  length  equal  to  the  slant  hight  of  the  frus- 
tum, as  shown  by  S*  T1  of  the  section.  Througli  W 
draw  N  II  indefinitely,  parallel  to  K  0.  At  right 
angles  to  K  0,  through  the  points  K  and  0,  draw 
lines  K  N  and  0  H,  cutting  N  II  in  the  points  N  and 
II,  thus  establishing  its  length.  Connect  M  N  and 


R  II.     Then   M  R  II  N  will  be  the  pattern  of  one  of 
the  four  sides  composing  the  article. 


PATTERN. 


Fig.  367. — The  Envelope  of  the  Frustum  of  a  Pyramid  Which  is 
Diamond  Shape  in  Plan, 


PROBLEM   76. 


The  Pattern  of  the  Flaring:  End  of  an  Oblong  Tub. 


In  Fig.  368,  A  B  D  C  shows  the  elevation  and 
N  P  O  R  the  plan  of  a  vessel  having  straight  sides 
and  semicircular  ends,  one  end  of  which  is  slanting. 
First  draw  a  correct  plan  and  elevation  of  the  article, 
seeing  that  each  point  of  the  elevation  is  carefully 
projected  from  its  corresponding  point  in  the  plan. 
Divide  half  of  the  boundary  line  of  the  top  into  any 
number  of  equal  spaces,  commencing  at  O,  all  as 
shown  by  the  small  figures  1,  2,  3,  etc.,  in  the  plan. 
From  the  points  thus  obtained  carry  lines  vertically 
until  they  cut  the  top  of  the  elevation,  as  shown  in 
the  points  between  B  and  L ;  also  continue  the  lines 
downward  until  they  meet  the  line  T  0,  all  as  shown. 
From  the  points  between  L  and  B  thus  obtained  draw 


lines  parallel  to  B  D,  producing  them  upward  indefi- 
nitely, and  continue  them  downward  until  they  meet 
the  bottom  line  of  the  elevation  F  D,  as  shown.  At 
right  angles  to  the  lines  thus  drawn,  and  at  any  con- 
venient distance  from  the  elevation,  draw  G  II.  With 
the  dividers,  set  off  from  the  line  G  H,  on  each  of  the 
lines  drawn  through  it,  the  distance  from  T  O,  on  the 
lines  of  corresponding  number,  to  the  curved  line  P  0. 
In  other  words,  make  G  K  equal  to  T  6  of  the  plan. 
Set  off  spaces  on  the  other  lines  corresponding  to  the 
distance  on  like  lines  in  the  plan.  Through  the  points 
thus  obtained  trace  a  line,  as  shown  by  K  H.  Then 
G  H  K  will  be  the  half  profile  of  the  end  of  the 
vessel  on  any  line,  as  L  M,  drawn  at  right  angles  to 


Pattern  Problems. 


179 


the  line  D  B.  The  stretchout  of  the  pattern  is  to 
be  taken  from  the  profile  thus  constructed.  At 
right  angles  to  D  II,  and  at  any  convenient  distance 
from  it,  draw  U  V,  upon  which  lay  off  twice  the 
stretchout  of  K  II,  numbering  each  way  from  the 
central  point  1,  as  shown.  From  the  points  in  the 
stretchout  thus  obtained  draw  measuring  lines  at  right 


spending  measuring  lines,  as  shown,  for  the  top  of  the 
pattern.  If  it  is  desired  to  make  a  joint  upon  the  line 
E  F  of  the  elevation,  the  triangular  shaped  piece  ELF 
may  be  added  to  the  pattern  as  follows  :  With  the  di- 
viders take  the  distance  E  F  of  the  elevation  as  radius, 
and  from  the  point  Z  of  the  pattern  as  center  describe  an 
arc.  In  like  manner,  with  E  L  of  the  elevation  as 


PLAN     R 
Fig.  368.— Pattern  of  the  Flaring  End  of  an  Oblong  Tub. 


angles  to  it  indefinitely.  With  the  blade  of  the 
T-square  set  at  right  angles  with  D  B,  and  brought 
successively  against  the  points  in  F  D,  cut  lines  of 
corresponding  number  drawn  through  the  stretchout. 
Then  a  line  traced  through  the  points  of  intersection, 
as  shown  by  Y  Z,  will  be  the  pattern  of  the  bottom 
end  of  the  piece.  In  like  manner  bring  the  blade  of 
the  T-square  against  the  points  in  L  B  and  cut  corre- 


radius,  and  point  R  in  the  upper  line  of  the  pattern  as 
center,  describe  a  second  arc,  cutting  the  first  arc  in 
the  point  W.  Connect  W  with  R  and  also  with  Z. 
The  triangular  piece  at  the  opposite  end  terminating 
in  point  X  is  added  in  a  similar  manner,  thus  com- 
pleting the  entire  pattern  of  that  portion  of  the  vessel 
from  the  line  E  F  to  the  right. 


PROBLEM    77. 


Pattern  for  the  Flaring  Section  of  a  Locomotive  Boiler. 


While  the  pattern  here  described  is  especially 
adapted  to  the  tapering  section  or  "  taper  course  "  of 
a  locomotive  boiler  its  principles  are  equally  applicable 
to  tanks,  cans  or  pipes  whose  shapes  are  governed  by 
the  spaces  or  positions  which  they  are  to  occupy.  The 


section  of  the  boiler  at  A  F,  Fig.  369,  is  round,  as 
shown  by  I  N  L  M.  The  lower  half  of  the  circle  I  M 
L  is  the  profile  from  L  F  to  G  D,  but  the  upper  half 
is  raked  or  slanted  from  B  K  to  C  II,  retaining  its  semi- 
circular character  at  C  H.  The  line  H  G  is  a  vertical 


180 


New  Metal   Wwkvr   /'«//>  m    liwk. 


line,  as  shown  by  S  L  of  the  sectional  view,  and  the 
surface  H  K  G  being  vertical  is  simply  a  flat  triangular 
surface,  exactly  as  shown  in  the  elevation. 


Fig.  369. — Pattern  for  the  Flaring  Section  of  a  Locomotive  Boiler. 

Since  the  part  B  K  H  C  is  semicircular  when  cut 
upon  any  vertical  line,  the  first  step  will  be  to  obtain 
a  section  of  it  as  it  would  appear  if  cut  upon  a  line  at 
right  angles  to  B  C.  This  section  must  be  derived 
from  the  normal  section  of  the  level  part,  and  may  be 
done  as  follows :  Assume  any  line,  as  U  ~W,  drawn  at 


right  angles  to  B  C  at  any  convenient  position  outside 
of  the  elevation,  us  the  vertical  center  lino  of  the  new 
section.  Divide  one-half,  the  normal  section,  as  N  L, 
into  any  convenient  number  of  spaces,  as  shown  bv  the 
figures,  and  from  the  points  thus;  obtained  draw  lines 
parallel  to  A  B,  cutting  B  K,  as  shown,  also  extend- 
ing them  back  to  the  center  line  N  M.  From  15  K 
carry  them  parallel  to  B  C,  cutting  the  line  C  II,  and 
extend  them  indefinitely,  cutting  also  the  line  U  AY. 
AVith  the  dividers  measure  the  horizontal  distance  of 
the  various  points  in  the  normal  profile  K  L  from  the 
center  line  N  M,  and  transfer  these  distances  to  lines 
of  corresponding  number,  measuring  each  time  from 
U  AY.  Thus  make  AA"  7  equal  to  O  7;  the  distance 
from  U  W  to  the  point  6  equal  to  P  l> ;  and  so  con- 
tinue till  all  the  distances  have  been  measured.  A 
line  traced  through  these  points  will  constitute  a  profile 

of  the  raking  portion  on  a  line 
at  right  angles  to  its  direction, 
undBK  and  C  il  will  be  the 
miter  lines.  To  develop  the 
pattern  first  layoff  the  stretch- 
out of  the  profile  T  U  Ar  upon 
any  line  drawn  at  right  angles 
to  B  C,  as  A'  B1.  As  the 
points  in  U  T  have  already 
been  dropped  upon  the  miter 
lines  in  the  previous  process 

it  is  now  only  necessary  to  place  the  T-square  parallel 
to  A1  B1  and,  bringing  it  successively  against  the 
points  in  C  II  and  B  K,  drop  lines  cutting  the 
measuring  lines  of  corresponding  number.  A  line 
traced  through  the  points  of  intersection,  as  shown  1>\- 
X  X  Y  Y,  will  be  the  pattern  of  the  raking  portion 
B  K  H  C.  To  this  may  be  added  the  flat  triangular 
piece  K  H  G,  as  shown  by  X  Y  Z.  From  the  points 
X  and  Z  lines  may  be  drawn  at  right  angles  to  X  /, 
as  shown  by  Z  J  and  X  Q,  extending  them  sufficiently 
to  complete  the  lower  portion  of  this  part  of  the  boiler, 
shown  by  K  E  D  G  of  the  elevation. 


PROBLEM  78. 

The  Pattern  for  a  Blower  for   a  Grate. 


In  Fig.  370  D  F  K  H  E  shows  a  front  view,  P  L 
M  0  a  side  view,  and  A  C  B  a  plan  of  a  blower.  The 
conditions  which  determine  the  course  to  be  pursued 
in  arriving  at  the  pattern  are  that  its  upper  outline 


shall  conform  to  the  semicircle  F  K  IT  of  the  eleva- 
tion, and  that  it  shall  slant  from  K  to.(i  at  an  angle 
indicated  by  the  line  L  M  of  the  profile.  Therefore, 
the  first  step  will  be  to  determine  a  true  section  of  its 


Pattern  Problems. 


181 


upper  portion  or  hood   upon  a  line  at  right  angles  to 
]j  M  from  the  puiut  N  of  the  side  view;  after  which, 


Fig.  370—  Pattern  for  a  Blower  for  a  Orate. 

with  N  M  and  N  L  as  miter  lines,  the  pattern  can  be 
developed  in  the  usual  manner.     To  obtain  a  true  pro- 


file of  the  hood,  divide  one-half  of  the  semicircle  F  K 
H  into  any  number  of  equal  spaces,  and  carry  lines 
from  each  of  the  several  points  to  the  vertical  line  L 
N  of  the  side  view.  From  the  points  thus  obtained  in 
L  N"  carry  lines  parallel  with  L  M  indefinitely,  which 
intersect  at  right  angles  by  the  line  T  S,  located  at  any 
convenient  point  outside  of  the  diagram.  With  the 
dividers  take  the  horizontal  distances  between  the  points 
in  the  arc  F  K  to  the  line  K  G,  and  set  them  off  on  the 
lines  of  corresponding  number,  measuring  from  the 
line  T  S.  Then  a  line  drawn  through  the  points  thus 
obtained,  and  as  indicated  by  T  R,  will  be  a  correct 
section  through  the  inclined  portion  of  the  blower. 
Take  the  stretchout  of  the  profile  T  R  point  by  point 
and  place  the  spaces  on  the  line  U  V,  which  is  drawn 
at  right  angles  to  L  M.  Through  the  points  in  U  V 
draw  the  usual  measuring  lines  at  right  angles  to  it. 
As  the  points  from  the  profile  T  R  have  already  been 
dropped  upon  both  the  miter  lines  M  N  and  N  L,  it  is 
only  necessary  to  carry  them,  at  right  angles  to  L  M, 
on  to  the  measuring  lines  of  corresponding  numbers. 
Then  a  line  traced  through  the  points  thus  obtained, 
and  as  indicated  by  I"  K'  H',  will  be  the  desired  pattern. 


PROBLEM  79. 


Pattern  for  a  Can  Boss  to  Fit  Around  a  Faucet. 


In  Fig.  371  is  shown  a  top  and  side  view  of  a 
boss  whoso  sides  are  in  part  parallel  and  just  suffi- 
ciently apart  to  allow  the  faucet  to  fit  between  them. 
L  N  represents  the  diameter  of  the  opening  at  the  top. 
K  L  M  X  represents  the  general  shape  of  the  boss 
where  it  joins  the  can  and  is  the  result  of  the  condi- 
tions existing  in  the  side  view,  but  is  not  made  use 
of  in  the  process  of  obtaining  the  pattern.  The  essen- 
tial points  are  the  curve  of  the  can  body,  D  A  B  E, 
the  diameter  of  boss  at  top,  L  0  N,  the  distance  be- 
tween 1)  and  E  and  the  distance  X  C,  all  of  which  are 
shown  in  the  side  view. 

Divide  one-quarter  of  the  plan  of  the  top,  as  indi- 
cated by  0  N,  into  any  convenient  number  of  spaces, 


as  indicated  by  1,  2,  3,  etc.  From  the  points  thus 
established  drop  lines  vertically,  cutting  the  line  repre- 
senting the  top  in  the  side  view,  as  shown  from  F  to 
C.  From  the  points  thus  established  in  F  C  carry 
lines  parallel  to  the  side  F  D,  producing  them  until 
they  cut  the  curved  line  D  A  B  E,  as  shown  between 
D  and  A.  The  next  step  to  be  taken  is  to  obtain  the 
profile  which  would  be  shown  by  a  section  taken 
through  the  article  at  right  angles  to  the  line  D  F. 
For  this  purpose  at  any  convenient  point  draw  a  line 
through  D  F  and  at  right  angles  to  it,  as  shown  by  P  R. 
From  the  points  established  in  the  plan  of  the  top,  as 
shown  from  O  to  N,  carry  lines  vertically  until  they 
meet  the  horizontal  line  K  M  passing  through  the  cen- 


182 


Neiv  Metal   Worker  Pattern  Book. 


ter  of  the  top,  as  shown.  Taking  the  length  of  each 
of  the  distances  thus  obtained  in  the  dividers,  set  it 
off  from  either  side  of  P  R  on  the  lines  of  correspond- 
ing numbers,  and  through  the  points  thus  obtained 
trace  the  curve,  as  shown.  Then  this  curve  will  repre- 
sent the  required  section  from  which  the  stretchout  of 
the.  envelope  may  be  obtained.  On  the  line  E  P,  pro- 
duced sufficiently  outside  of  the  side  view  for  the  pur- 
pose, lay  off  the  stretchout  of  one-half  of  this  curve, 
as  shown,  and  through  the  points  thus  established 
draw  measuring  lines  parallel  to  D  F.  Then,  with 
the  T-square  placed  parallel  to  P  R,  or,  what  is  the 
same,  at  right  angles  to  D  F,  and  brought  successively 
against  the  points  in  the  profile  of  the  can  body  be- 
tween D  and  A,  cut  the  measuring  lines  of  correspond- 
ing numbers.  In  like  manner  bring  the  J-square 
against  the  points  in  the  top  of  the  article  shown  from 
F  to  C  and  cut  the  measuring  lines  of  corresponding 
numbers.  Then  lines  traced  through  the  points  thus 
obtained,  as  shown  by  D'  A'  and  F'  C',  will  be  one- 
half  of  the  pattern  of  one  of  the  ends.  As  that  por- 
tion of  the  boss  lying  between  points  A,  B  and  C  is 
simply  a  flat  triangular  piece  it  is  only  necessary  to 
add  a  duplicate  of  its  shape  to  that  part  of  the  pattern 
just  obtained,  bringing  one  of  its  straight  sides  against 
the  line  5,  all  as  shown.  To  the  other  straight  side 
C'  B'  must  be  added  a  duplicate  of  the  first  part  of  the 
pattern  reversed,  as  shown  by  B'  C'  G'  E';  the  result- 
ing shape  will  then  constitute  the  pattern  of  one-half 
the  boss. 


Fig.  371.— Pattern  for  a  Can  Boss  to  Fit  Around  a  Faucet. 


PROBLEM  80. 


The  Patterns  for  a  Molded  Base  in  which  the  Projection  of  the  Sides  is  Different  from  that  of  the  Ends. 


Let  A  B  C  D,  in  Fig.  372,  represent  the  side  view 
and  E  F  G  H  the  plan  of  a  base  in  which  the  projec- 
tion of  the  sides,  as  shown  at  O  P,  is  less  than  that  of 
the  ends,  as  shown  at  M  C.  B  C  and  A  D  show  the 
profile  of  the  ends  of  the  base.  As  the  projection 
through  the  sides  of  the  base  is  less  than  that  of  the 
ends,  a  profile  must  be  obtained  through  the  side  0  P 
in  plan,  from  which  to  obtain  the  stretchout  in  pro- 
ducing the  pattern  of  sides.  To  obtain  the  pattern  for 
the  end  proceed  as  follows :  Divide  the  profile  B  C 
into  an  equal  number  of  parts,  as  shown  by  the  small 
figures,  and  from  the  points  obtained  drop  lines  at  right 


angles  to  A  B  until  they  intersect  the  miter  lines  J  P 
and  L  Gr  in  plan.  At  right  angles  to  F  G  draw  the  line 
B"  C",  upon  which  place  the  stretchout  of  the  profile 
B  C,  as  shown  by  the  small  figures,  through  which  draw 
the  usual  measuring  lines,  which  intersect  with  lines 
of  corresponding  numbers  drawn  from  the  miter  lines 
at  right  angles  to  F  G.  A  line  traced  through  these 
intersections,  as  shown  by  J'  L'  G'  F',  will  be  the  re- 
quired pattern  for  the  end  of  the  base.  To  obtain  the 
profile  through  the  side  proceed  as  follows :  From  B 
in  elevation  draw  the  vertical  line  B  M,  as  shown,  and 
from  the  divisions  on  the  profile  B  C  draw  lines  par- 


Pattern    Problenifi. 


183 


allel  to  the  lines  of  the  moldings  until  they  intersect 
the  line  B  M,  as  shown  l>v  tlie  small  figures.  From 
the  intersections  obtained  on  the  miter  line  L  G  in  plan, 
as  before  explained,  draw  lines  parallel  to  H  G  and  ex- 
tend them  indefinitely,  cutting  the  miter  line  K  H,  as 
shown.  Upon  K  L  of  the  plan  extended  set  off  IV  M', 


the  side  of  the  base.  To  obtain  the  pattern  for  the 
side  proceed  as  follows  :  At  right  angles  to  II  G  draw 
the  line  B*  N",  upon  which  place  the  stretchout  of  the 
profile  B'  N',  as  shown  by  the  small  figures.  It  will  be 
noticed  that  the  spaces  in  the  profile  B'  N'  are  unequal, 
and  therefore  each  must  be  separately  placed  on  the  line 


N'        W 

PROFILE  THROUGH 
0.  P.  IN"  PLAN, 


F'      , 


Fig. 


H'  N2  G 

Patterns  for  a  Molded  Base  in  Which  the  Projection  of  the  Sides  {s  Different  from  that  of  the  Ends. 


in  hight  equal  to  B  M  of  elevation,  and  transfer  the 
spaces  from  B  M  to  B'  M',  as  shown.  At  right  angles 
to  B'  M'  and  from  the  points  on  same  draw  lines,  as 
shown,  intersecting  lines  of  corresponding  numbers 
drawn  from  the  rniter  line  L  G.  A  line  traced  through 
these  intersections  will  be  the  desired  profile  through 


B8  NJ.  Through  the  points  in  this  stretchout  line  draw 
the  usual  measuring  lines,  as  shown,  which  intersect 
with  lines  of  corresponding  numbers  drawn  at  right 
angles  to  H  G  from  the  miter  lines  K  II  and  L  G.  A 
line  traced  through  these  intersections,  as  shown  bv 
K'  H'  G'  L',  will  be  the  desired  pattern  for  the  side. 


PROBLEM    81. 

The  Patterns  for  an  Elliptical  Vase  Constructed  in  Twelve  Pieces. 


The  first  essential  in  beginning  the  work  is  an 
ellipse,  which  may  be  drawn  by  whatever  rule  is  most 
convenient,  and  which  must  be  of  the  length  and 
breadth  which  the  vase  is  required  to  have.  Draw  the 
plan  of  the  sides  of  the  vase  about  the  curve,  as  shown 


in  Fig.  373,  in  such  a  manner  that  all  the  points  X,  Y, 
Z,  etc.,  shall  have  the  same  projection  beyond  the 
curve.  Complete  at  least  one-fourth  of  the  plan  by 
drawing  miter  lines,  as  shown  by  P  C,  M  C,  0  C,  U  C 
and  N  C.  Above  the  plan  construct  an  elevation  of 


184 


Tin1   \/'/r   Mini    II '<;/•/•>•;•   I'lilli-ni 


the   article,  as  shown  bv  II  L  K  G.      Only  the  profile      considered  before  the  article    is  constructed.      As  the 
H  V  W  L  of  the  elevation  is  needed  for  the  purpose  of      projection  of  cadi  of  lhe  sides  upon  the  plan  is  differ- 


Fig.  37S.—The  Patterns  for  an  Elliptical  Vase  Constructed  in  Twelve  Puves. 


pattern    cutting,  but  the   other  lines  are  desirable  in 
process  of  designing,  in  order  that  the  effect  may  be 


ent  when  measured  from  the  center  C  on  lines  at  right 
angles  to  the  lines  of  the  sides,  it  will  be  necessary 


Pattern 


185 


lirst  to  develop  the  profiles  of  each  of  the  van  ing  sides 
from  the  normal  prolile  II  V  \V  L.  Therefore.  di\  idc 
-II  Y  \\'  Ij  in  the  usual  manner,  and  from  the  several 
points  in  it  drop  lines  across  its  corresponding  section 
(.No.  1)  of  the  plan. 

Across  the  second  section  in  the  plan,  from  the 
points  already  obtained  in  U  C,  draw  lines  parallel  to 
<)  U,  the  side  of  it  cutting  O  C,  and  produce  them 
until  thcv  meet  A  C,  which  is  drawn  from  C  at  right 
angles  to  1'  <)  produced.  Then  the  points  in  AC  give 
the  projections  from  which  to  obtain  a  profile  of  the 
section  numbered  '2.  In  like  manner  continue  the 
points  from  C  O  across  the  third  section  in  the  plan, 
parallel  to  O  M,  the  side  of  it  cutting  M  C,  and  pro- 
duce them  until  they  cut  C  15,  which  is  drawn  from  C 
at  right  angles  to  O  M  produced.  Then  C  B  contains 
the  points  requisite  in  obtaining  a  profile  of  the  third 
section.  Continue  the  points  in  C  M  across  the  fourth 
section,  cutting  its  other  miter  line  C  P.  From  C 
draw  C  D  at  right  angles  to  the  side  P  M  of  the  sec- 
tion, cutting  the  lines  drawn  across  section  4.  Then 
upon  C  D  will  be  found  the  points  necessary  to 
determine  the  profile  of  the  fourth  pattern.  Pro- 
duce the  line  of  the  base  of  the  elevation  indefinitely, 
as  shown  by  C'  C"  C:l.  and  also  the  line  of  the  top  A' 
W  I)'.  From  the  several  points  in  the  profile  H  Y  "W 
L  draw  lines  indefinitely,  parallel  to  the  lines  just  de- 
scribed and  as  shown  in  the  diagram.  From  C1,  upon 
the  base  line  produced,  set  off  points  corresponding  to 


the  points  in  C  A  of  the  plan,  making  the  distance 
from  C'  in  each  instance  the  same  as  the  distance  from 
C  in  the  plan.  NumVr  the  points  to  correspond  with 
the  numbers  given  to  the  points  in  the  profile  II  V  W 
L,  from  which  they  were  derived.  In  like  manner 
from  C*  set  off  points  corresponding  to  the  points  in  C 
B  of  the  plan,  numbering  them  as  above  described. 
From  C3  set  off  points  corresponding  to  those  in  CD  of 
the  plan,  likewise  identifying  them  by  figures  in  order 
to  facilitate  the  next  operation.  From  C1  erect  the 
perpendicular  C'  A1 ;  likewise  from  C"  and  C3  erect  the 
perpendiculars  C2  B°  and  C3  D3.  From  each  of  the 
points  laid  off  from  C1,  and  also  from  each  of  those  laid 
off  from  C'  and  C3,  erect  a  perpendicular,  producing  it 
until  it  meets  the  horizontal  line  drawn  from  the  profile 
H  V  W  L  of  corresponding  number.  Then  lines  traced 
through  these  several  intersections  will  complete  the 
profiles,  as  shown.  Perpendicular  to  the  side  of  each 
section  in  the  plan  lay  off  a  stretchout  taken  from  the 
profile  corresponding  to  it,  just  described,  and  through 
the  points  in  the  stretchouts  draw  measuring  lines  in 
the  usual  manner,  all  as  shown  by  E  F,  E'  F',  EJ  FJ  and 
E3  F3.  Place  the  T-square  parallel  to  each  of  these 
stretchout  lines  in  turn,  and,  bringing  it  against  the 
several  points  in' the  miter  lines  bounding  the  sections 
of  the  plan  to  which  they  correspond,  cut  the  measur- 
ing lines  in  the  usual  manner.  Then  lines  tracad 
through  the  points  of  intersection  thus  obtained,  all  as 
shown  in  the  diagram,  will  complete  the  patterns. 


PROBLEM  82. 


The  Patterns  for  a  Finial,  the  Plan  of  which  is  an  Irnegfular  Polygon. 


In  the  central  portion  of  Fig.  374  is  shown  the 
plan  B  C  D  E  F  G  II,  upon  which  it  is  required  to 
construct  a  fiuial,  the  only  other  view  given  being  a 
section  through  one  of  the  sides,  being  that  numbered 
1  on  the  plan. 

The  section  of  side  ABC,  or  No.  1,  is  shown 
above  and  in  line  with  the  plan  of  the  same,  and  is 
marked  Profile  No.  1,  and  is  a  section  on  the  line  A  M, 
which  is  drawn  at  right  angles  to  B  C  of  the  plan. 
To  obtain  the  pattern  of  A  B  C  of  plan,  or  No.  1, 
divide  the  profile  K  L  in  the  usual  manner,  and,  with 


the  T-square  placed  parallel  with  C  B  of  plan  and 
brought  successively  against  the  several  points  in 
profile  K  L,  drop ;  lines  cutting  the  miter  lines  A  B 
and  A  C. 

On  A  M  extended,  as  Al  Ml,  lay  off  a  stretchout 
.of  K  L  of  profile,  through  the  points  in  which  draw 
the  customary  measuring  'lines.  Place  the  T-square 
parallel  to  the  stretchout  line  Al  Ml,  and,  bringing  it 
against  the  several  points  in  the  miter  lines  A  B  and 
A  C,  cut  corresponding  measuring  lines.  Tracing 
lines  through  the  points  thus  obtained,  as  shown  by 


186 


The  New  Metal   Worker  Pattern  Book. 


Al  Cl  Bl,  will   give   the    pattern  of    part  of    article 
shown  on  plan  by  A  B  C. 


at  right  angles  to  eacli — that  is,  since  A  N,  A  0  and 
A  P,  drawn  at  right  angles  respectively  to  C  D,  D  E 


PROFILE 
NO.4 


PROFILE 
NO. 3 


PROFILE 
NO.1 


K2 


PROFILE 
NO.2 


cij 


D3 


02 


C2 


Fig.  374.— Pattern  for  a  F inial,  the  Plan  of  Which  is  an  Irregular  Polygon. 


Since  the  point  A  in  the  plan  is  not  equidistant 
from  all  the  sides  of  the  same,  when  measured  on  lines 


and  E  F,  differ  in   length   from  each   other  and  from 
A  M a  correct  section  must   be   obtained  for   each 


Pattern   Problems, 


187 


of  the  other  sides  before  their  patterns  can  be 
developed. 

These  different  sections  can  be  most  conveniently 
obtained  at  one  operation  in  the  following  manner: 

With  the  T-square  placed  parallel  with  D  C,  and 
brought  successively  against  the  points  in  A  C,  draw 
lines  cutting  A  D  and  A  N.  Then  the  points  in  A  N 
can  be  used  to  obtain  a  profile  of  section  No.  2.  Also 
continue  the  points  from  A  D  across  the  third  section 
of  the  plan,  and  parallel  with  D  E,  and  produce  them 
until  they  cut  A  E,  also  A  0,  which  is  drawn  from  A 
at  right  angles  to  D  0  produced.  Then  A  0  contains 
the  points  necessary  in  obtaining  a  profile  of  the  third 
section.  Continue  the  points  from  A  E  across  the 
fourth  section  of  the  plan,  and  parallel  with  E  F,  cut- 
ting A  P  and  A  F.  Then  the  points  in  A  P  can  be 
used  to  obtain  a  profile  of  section  No.  4. 

While  the  projection  of  the  several  points  in  each 
of  the  new  profiles  can  be  obtained  respectively  from 
the  lines  A  N,  A  0  and  A  P,  the  hights  of  the  several 
points  must  be  the  same  in  all  and  must  be  derived 
from  the  normal  profile  K  L.  Therefore  continue 
J  L,  which  represents  a  horizontal  line  of  profile  No.  1, 
in  either  direction,  as  shown  by  L4  L2.  From  the 
several  points  in  profile  No.  1  draw  lines  parallel  with 
L4  L2,  extending  them  indefinitely  in  either  direction. 
At  any  convenient  position  on  L4  L2  set  off  points 
corresponding  to  the  points  in  A  N  of  the  plan,  as 
shown  at  J2  L2,  numbering  them  to  correspond  with 


the  points  in  A  N  and  in  the  normal  profile.  From 
each  of  the  points  in  J2  L2  erect  lines  perpendicular 
to  the  same,  intersecting  lines  of  corresponding  number 
drawn  from  K  L.  Then  a  line  traced  through  the  points 
of  intersection,  as  shown  from  K2  to  L2,  will  be  the  cor- 
rect section  on  A  N  of  the  plan,  from  which  to  obtain 
a  stretchout  of  piece  No.  2. 

The  sections  of  pieces  No.  3  and  No.  4  are  ob- 
tained in  a  similar  manner  from  A  O  and  A  P.  J3 
L3  is  a  duplicate  of  A  0  and  J4  34  of  A  P.  Perpen- 
diculars are  erected  from  each  of  the  points  cutting 
horizontal  lines  of  corresponding  number,  thus  devel- 
oping K3  L3  and  K4  L4. 

To  obtain  the  pattern  of  A  D  C  (No.  2)  continue 
A  N  downward  indefinitely,  upon  which  lay  off  a 
stretchout  of  K2  L2,  as  shown  by  A2  N2,  through 
the  points  in  which  draw  the  usual  measuring  lines. 
Place  the  "[-square  parallel  with  A2  N2,  and,  bringing 
it  against  the  several  points  in  A  C  and  A  D,  cut 
measuring  lines  of  corresponding  numbers.  Lines 
traced  through  these  points,  as  shown  by  A2  C2  and 
by  A2  D2,  will  be  the  pattern  sought. 

The  stretchouts  for  pieces  Nos.  3  and  4  are  taken 
respectively  from  profiles  3  and  4.  A3  03  is  the 
stretchout  of  K3  L3  and  is  laid  off  on  a  continuation 
of  A  O,  while  A4  P4  is  taken  from  K4  L4  and  is  set 
off  on  a  continuation  of  A  P.  The  remaining  opera- 
tions are  the  same  as  those  employed  in  obtaining  the 
other  pieces. 


PROBLEM  83. 


Pattern   for  a  Three-Piece  Elbow,  the  Middle  Piece  Being:  a  Gore. 


Let  A  B  C  D  E  F  Gr  in  Fig.  375  be  the  elevation 
of  a  three-piece  elbow  to  any  given  angle,  as  G  F  E, 
the  middle  piece  of  which,  B  C II,  forms  a  gore  extend- 
ing around  one-half  the  diameter.  The  lines  H  B  and 
II  C  are  drawn  parallel  respectively  to  the  ends  of  the 
two  outer  pieces,  therefore  the  patterns  for  the  end 
pieces  will  be  straight  from  H  to  C  and  H  to  B  and 
mitered  from  H  to  F.  To  obtain  the  pattern  for  one 
of  the  ends,  as  F  H  ODE,  divide  N  K  L  into  any 
convenient  number  of  equal  parts.  With  the  7-square 
at  right  angles  to  N  L,  carry  lines  from  the  points  in 


N  K  L,  cutting  the  miter  line  F  H  C,  as  shown.  (X 
any  line,  as  E  D  extended,  lay  off  a  stretchout  of  M 
K  L,  as  shown  by  e  d,  through  the  points  in  which 
draw  the  usual  measuring  lines,  as  shown.  With  the 
T-square  brought  successively  against  the  points  in  F 
II  C,  cut  corresponding  measuring  lines,  as  shown.  A 
line  traced  through  the  points  of  intersection,  as  shown 
by  fh"  c,  will  be  the  half  pattern  of  end,  as  represented 
in  elevation  by  F  H  C  D  E,  or  in  profile  by  N  K  L. 
The  other  half  of  the  pattern  can  be  obtained  by  dupli- 
cation. 


188 


The  New  Metal   HW.vr  Pattern  Bool'. 


Since  II  C  is  drawn  parallel  to  E  D,  the  distance 
II  J  is  less  than  one-half  the  diameter  of  the  normal 
profile,  or  less  than  0  L,  therefore  it  will  lie  necessary 
before  obtaining  the  pattern  of  B  II  C  to  obtain  a  cor- 
rect section  on  line  H  J  of  elevation.  To  do  this, 
first  place  tlie  T-square  parallel  with  B  C  and  carry  lines 
from  the  points  in  H  C,  cutting  II  J  and  H  B.  Next 
draw  any  line,  as  K'  M'  of  the  section,  and  erect  the 
perpendicular  H'  J'.  From  II',  on  H'  J',  set  off  the 
spaces  in  H  J  of  elevation,  transferring  them  point  by 
point.  Through  the  points  thus  obtained  draw  lines 
parallel  with  K'  M',  as  shown.  With  the  dividers  take 
the  distance  across  K  O  L  or  L  O  M,  on  the  several 
lines  drawn  parallel  with  K  M,  and  set  off  the  same 
distance  on  lines  of  corresponding  number  drawn 
through  H'  J'.  Thus  H'  M'  and  H'  K'  are  the  same  as 
0  K  and  0  M.  A.  line  traced  through  the  points  thus 
obtained,  as  shown  by  K'  J'  M',  will  be  the  section 
desired. 

For  the  pattern  of  II B  C,  lay  off  on  H  J  extended, 
as  h  h',  a  stretchout  of  K'  J'  M'  of  section,  through  which 
draw  the  usual  measuring  lines.  With  the  T-square 
placed  parallel  with  II  J,  and  brought  successively 
against  the  points  in  B  II  and  C  H,  cut  corresponding 
measuring  lines  drawn  through  k  h',  as  indicated  by  the 
dotted  lines.  Lines  drawn  through  these  points  of 
intersection,  as  shown  by  1>  1>  h'  r,  will  be  the  pattern 
of  the  part  shown  in.  elevation  by  B  II  0. 


K' 


H' 
SECTION. 


Fig.  37S.—lttltei-n  /../•  a  Three-Piece  Elbuw,  tlie  Middli-  Piece 
lie  in  ij  a  (Jure. 


PROBLEM  84. 


The  Patterns  of  a  Tapering  Article  which  is  Square  at  the  Base  and  Octagonal  at  the  Top. 


A  B  D  C  in  Fig.  376  shows  the  plan  of  the  article 
at  the  base,  IKLMHGrFE  represents  the  shape  at 
the  top,  E3  H'J  D"  C3  is  an  elevation  of  one  side.  In 
order  to  obtain  the  slant  hight  of  the  octagonal  sides  it 
will  be  necessary  to  construct  a  diagonal  section  or  ele- 
vation. Therefore  extend  the  lines  of  the  base  C3 1)2  and 
top  E3ir,  as  shown,  to  the  left,  through  which  draw  a 
vertical  line,  as  R1  C2.  Upon  the  line  of  the  base  set 
off  each  way  from  C'  a  distance  equal  to  A  R  or  R  1) 
of  the  plan.  In  like  manner  upon  the  line  of  the  top, 
set  off  from  R'  each  way  a  distance  equal  to  one-half 


I  G.  Draw  I1  A'  and  G'  D1,  thus  completing  a  diagonal 
section.  If  it  is  desired  to  complete  a  diagonal  eleva- 
tion, set  off  Es  F"  equal  to  E  F  of  the  plan  and  draw 
lines  to  C!,  as  shown. 

To  obtain  the  pattern  of  one  of  the  smaller  sides, 
produce  the  diagonal  line  R  C,  upon  which  set  off  the 
length  or  stretchout  of  I'  A1,  as  shown  by  N  C',  and 
draw  the  measuring  line  E'  F1.  By  means  of  the 
T-square,  as  indicated  by  the  dotted  lines,  set  off  E1  K1 
equal  to  E  F  of  the  plan  and  draw  C'  E1  and  C'  F'. 
Then  E'  C'  F'  is  the  pattern  of  one  of  the  smaller  sides 


1'iitii  i-ii    Problems. 


189 


Fig.  377.— Pattern  in  One  Piece, 

of  the  article.  For  the  pattern  of  one  of  the  larger 
sides,  draw  11  P  perpendicular  to  the  side  A  C,  upon 
which  set  oil'  0  P,  in  length  equal  to  E3  C'  of  the  ele- 
vation, at  right  angles  to  which  through  0  and  P  draw 
measuring  lines. 

By  means  of  the  T-square,  as  shown  by  the  dotted 
lines,  make  A2  C4  equal  to  A  C  of  the  plan.  In  like 
manner  make  F  E'  equal  to  I  E  of  the  plan.  Connect 
A2  P  and  C4  E4.  Then  A'  T  E4  C4  will  be  the  pattern 
of  one  of  the  larger  sides  of  the  article.  If  for  any 
reason  the  pattern  is  desired  to  be  all  in  one  piece  the  , 
shapes  of  the  different  sides  may  be  laid  off  adjacent 
in  c:ich  other,  the  large  and  small  sides  alternating,  all 
as  indicated  by  i  £  a  a1,  Fig.  377. 


Pig.  376.— Elevations,  Plan  and  Patterns. 
A  Taperinij  Article  Which  is  Square  at  the  Base  and  Octagonal  at  the  Top. 


PROBLEM   85 

The  Patterns  of  a  Finial,  the  Plan  of  which  is  Octagon  with  Alternate  Long:  and  Short  Sides. 


In  Fig.  378,  let  A  L  M  N  0  P  K  S  T  be  the  ele- 
vation of  the  iinial  corresponding  to  the  plan  which  is 
shown  immediately  below  it.  The  elevation  is  so 
drawn  as  to  show  the  profile  of  one  of  the  long  sides, 
for  the  pattern  of  which  proceed  in  the  usual  manner. 
Divide  the  profile  A  L  M  N  O  P  into  any  number  of 
convenient  spaces,  as  shown  by  the  small  figures,  and 
from  the  points  thus  obtained  drop  lines  across  tin; 
corresponding  section  in  the  plan,  cutting  the  miter 
lines  DEC  and  D2  F  C,  as  shown.  A  duplicate  of  the 
part  E  C  F  of  the  plan  is  shown  below  by  K1  C2  F',  and 
in  the  demonstration  C2  K1  and  0"  K'  are  to  be  consid- 
ered the  same  in  all  respects  as  C  K  and  C  F.  The 
same  mav  be  said  of  the  two  parts  of  the  stretchout 
line  bearing  like  letters;  the  division  having  been 


made  on  account  of  the  extreme  length  which  the  pat- 
tern would  have  if  made  in  one  piece.  Perpendicular 
to  D  D2  lay  off  a  stretchout,  as  shown  by  G  H,  through 
the  points  in  which  draw  measuring  lines  in  the  usual 
manner.  Place  the  y-square  parallel  to  the  stretchout 
line,  and,  bringing  it  against  each  of  the  several  points 
in  D  E  C  and  D2  F  C,  cut  the  corresponding  measuring 
lines.  Then  a  line  traced  through  these  points  of 
intersection  will  be  the  pattern  sought. 

For  the  pattern  of  the  short  sides  a  somewhat  dif- 
ferent course  is  to  be  pursued.  As  the  distance  from 
C  to  the  line  K  E  is  greater  than  that  from  C  to  E  F  a 
profile  of  the  piece  as  it  would  appear  if  cut  on  the  line 
C  1)  must  first  be  obtained.  To  do  this  proceed  as 
follows :  From  the  points  in  C  E,  dropped  from  the 


190 


The  Sew  Metal    Worker  Pattern  Bwk. 


1 

31+7  4     '6 

B1  5     17 

Fig.  378.— Elevation,  Plan  and  Patterns.  Fig.  379. —Diagonal  Section. 

The  Patterns  of  a  Finial,  the  Plan  of  Which  is  Octagon  with  Alternate  Loruj  and  Short  Sides. 


profile,  carry  lines  parallel  to  E  K  across  C  D,  cutting 
C  K,  as  shown.  At  any  convenient  place  lay  off  B' 
JP1,  Fig.  379,  in  length  equal  to  C  D  of  the  plan,  on 


which  lay  off  points  corresponding  to  the  points  ob- 
tained in  C  D,  and  for  convenience  in  the  succeeding 
operations  number  them  to  correspond  with  the  num- 


Pattern   Problems. 


191 


Vrs  in  the  profile  from  whirh  they  are  derived.  At 
IV  erect  the  perpendicular  B'  A',  equal  to  B  A  of  the 
elevation.  From  the  several  points  in  the  profile  of 
the  elevation  draw  horizontal  lines  cutting  the  'central 
vertical  line  A  B,  as  shown.  Set  off  points  in  A'  B1  in 
Fig.  379  to  correspond,  and  through  these  points  draw 
horizontal  lines,  which  number  for  convenience  of  iden- 
tification in  the  following  steps.  From  the  several  points 
in  B1  P'  carry  lines  vertically,  intersecting  correspond- 
ing horizontal  lines.  Then  a  line  traced  through  these 
points,  as  shown  by  A'  L'  M'  N'  O1  P',  will  be  the 
profile  of  the  short  side  on  the  line  C  D  of  the  plan. 


After  obtaining  the  profile  as  here  described,  for  the 
pattern  of  the  short  side  proceed  as  follows :  Perpen- 
dicular to  K  E  of  the  short  side^  or  on  C  D  extended, 
lay  off  a  stretchout  of  the  diagonal  section  A1  O'  of 
Fig.  379,  as  shown  by  C'  D1,  through  the  points  in 
which  draw  measuring  lines  in  the  usual  manner.  Place 
the  T-square  parallel  to  the  stretchout  line,  and,  bring- 
ing it  against  the  several  points  in  the  miter  lines  D  K 
C  and  DEC  bounding  the  short  side  in  the  plan,  cut 
the  corresponding  measuring  lines.  Then  a  line  traced 
through  these  points,  as  shown  in  the  diagram,  will  be 
the  required  pattern. 


PROBLEM  86. 

The  Pattern  for  a  Gore  Piece  Forming  a  Transition  from  an  Octagon  to  a  Square,  as  at  the 

End  of  a  Chamfer. 


In  Fig.  380,  let  F  F  F  F  represent  the  plan  of  the 
square  portion  of  a  shaft  and  A  A  A  A  that  of  the 


Pattern 


Fig.  380.— The  Pattern  for  a  Gore  Piece  Forming  the  Transition 
from  an  Octayon  to  a  Square. 

octagon  portion.  Let  D  P  C  be  the  elevation  of  the 
gore  piece  which  is  required  to  form  the  transition  be- 
tween the  two  shapes.  The  outline  C  D,  which  repre- 


sents the  intersection  of  the  gore  piece  with  the  side 
of  the  shaft,  may  be  of  any  contour  whatever  at  the 
pleasure  of  the  designer,  the  method  of  laying  out  the 
pattern  being  the  same  no  matter  what  its  outline.  By 
reference  to  the  plan  it  will  be  seen  that  the  lines  of 
the  molding,  of  which  C  D  shows  only  the  termination, 
run  octogonally,  or  in  the  direction  of  A  A.  There- 
fore, before  a  stretchout  of  the  piece  can  be  obtained 
a  correct  profile  must  be  developed  on  a  line  at  right 
angles  to  its  lines — that  is,  on  the  line  E  F.  To  do  this 
proceed  as  follows :  Divide  the  line  C  D,  as  it  appears 
in  the  elevation,  into  any  convenient  number  of  spaces, 
as  shown  by  1,  2,  3,  4,  etc.  From  the  points  thus 
obtained  drop  lines  down  upon  the  side  of  the  plan 
A  F,  which  should  be  placed  in  line  below  the  ele- 
vation. 

Continue  the  lines  from  the  side  A  F  across  the 
corner,  as  shown,  all  parallel  to  A  A  of  the  octagon, 
crossing  E  F,  and  number  the  lines  to  correspond 
with  the  numbers  of  the  points  in  the  elevation  from 
which  they  were  derived.  Draw  the  vertical  line 
G  n  at  a  convenient  distance  from  P  D,  and  cut  G  n 
by  lines  drawn  at  right  angles  to  it  from  the  points  in 
C  D,  as  shown  by  the  connecting  dotted  lines.  G  II 
then  may  be  considered  to  represent  the  point  F  in  the 
plan,  or  11  of  the  numbers  on  the  line  E  F.  From 
G  II,  on  each  of  the  several  lines  drawn  through  it,  lay 
off  a  distance  equal  to  the  space  from  E  to  the  corre- 
sponding number  in  the  same  plan.  Thus  lay  off  from 
G  H  on  line  1  a  distance  equal  to  11  1  on  E  F,  and 
on  line  2  a  distance  equal  to  11  2  of  E  F,  and  so  on 
for  each  of  the  lines  through  G  H.  Then  a  line  traced ' 
through  these  points,  as  shown  by  I  H,  will  be  the 


192 


The     New  Metal    ll'w/ar  Pattern  Book. 


profile  of   the  gore  piece,   or  the  shape  of  its  section 
when  cut  by  the  line  E  F. 

Prolong  E  F,  as  shown  by  K  L,  and  lay  off  on 
the  latter  a  stretchout  of  the  profile  I  H,  the  spaces  of 
which  must  be  taken  from  point  to  point  as  they  occur, 
so  as  to  have  points  in  the  stretchout  corresponding  to 
the  points  on  the  miter  lines  A  F,  previously  derived 


from  C  D.  Through  the  points  thus  obtained  draw 
the  usual  measuring  lines,  as  shown.  Place  the 
T-square  at  right  angles  to  the  measuring  lines,  or,  what 
is  the  same,  parallel  to  E  F,  and,  bringing  it  against 
the  points  in  A  F  and  F  A,  cut  the  corresponding  lines 
drawn  through  the  stretchout.  Lines  traced  through 
these  points,  as  shown,  will  constitute  the  pattern. 


PROBLEM   87. 

The  Pattern  for  a  Gore  Piece  in  a  Molded  Article,  Forming:  a  Transition  from  a  Square  to  an  Octagon. 

In  Fig.  381,  let  A  B  D  C  represent  the  elevation 
of  an  article  of  which  G  H  I  J  is  the  half  plan  at  the 
base  and  K  L  M  N  O  P  the  half  plan  at  the  top.  A  C 
of  the  elevation  is  the  normal  profile  or  profile  of  one 
of  the  square  sides,  and  L  H  and  M  II  of  the  plan 
show  the  miter  lines  between  the  square  sides  and  the 
gore  piece.  C  E  and  D  F,  the  elevations  of  the  miter 
lines  II  M  and  I  N,  are  shown  as  part  of  the  design, 
but  are  not  necessary  in  cutting  the  pattern. 

As  only  the  normal  profile,  which  would  be  used 
in  cutting  the  pattern  of  one  of  the  square  sides,  is 
shown  in  the  elevation,  the  first  step  will  be  to  obtain 
from  this  a  profile  of  the  gore  piece,  or  in  other  words, 
a  section  upon  its  center  line,  R  H  of  the  plan.  Divide 
the  profile  A  C  into  any  convenient  number  of  parts, 
and  from  the  points  obtained  drop  lines  at  right  angles 
to  A  B,  cutting  the  miter  line  L  H  in  plan,  as  shown. 
From  the  intersections  obtained  on  the  miter  line  L  II 
draw  lines  parallel  to  L  M,  as  shown,  cutting  the  other 
miter  line  H  M,  and  continue  them  indefinitely.  At 
any  convenient  position  outside  the  plan  draw  the  line 
A1  A"  parallel  to  H  R,  and  draw  a  duplicate  of  the 
profile  A  C  in  the  same  relative  position  to  A1  A2  that 
A  C  holds  to  A  B,  and  divide  the  same  into  the  same 
spaces  as  A  C,  all  as  shown  by  A1  C'.  From  the  points  in 
A1  C1  draw  lines  parallel  to  A1  A",  cutting  lines  of 
corresponding  number  drawn  through  the  plan  of  the 
gore  piece.  A  line  traced  through  these  intersections,  as 
shown  from  A"  to  C",  will  be  the  profile  of  the  transition 
piece  from  which  to  obtain  the  stretchout  for  the  pattern. 

To  obtain  the  pattern  proceed  as  follows  :  Upon 
K'  R2,  a  continuation  of  II  R,  place  the  stretchout  of 
the  profile  A'  Ca,  as  shown  by  the  small  figures,  through 
which  draw  the  measuring  lines,  as  shown.  These  are 
n<>\v  to  be  intersected  bv  lines  drawn  from  points  of 
corresponding  number  upon  the  miter  lines  H  L  and 
II  M.  Lines  traced  through  the  points  of  intersection, 

Fig.  X81.—The  Pattern  for  a  Gore  Piece  in  a  Molded  Article,  Forniinrj 

shown  by  R  T  and  R  S,  will  give  the  desired  pattern.  a  Transition  from  a  Square  to  an  Octa,j,,lt. 


PROFILE  OF 
GORE  PIECE. 


f  '/!//<  r  n 


lit  3 


PROBLEM  88. 

The  Patterns  for  a  Raking:  Bracket. 


Tliis  is  one1  of  the  many  instances  which  calls  for 
special  draftsmanship  on  the  part  of  the  pattern  cutter. 
Frequently  the  architect's  drawings  give  only  a  detail 
of  a  bracket  for  the  level  cornice  of  a  building,  while 
the  scale  elevations  show  one-  or  more  of  the  gables  to 
be  finished  with  raking  brackets.  In  such  cases  the 
detail  of  the  "  level  "  bracket,  and  the  pitch  of  the  roof 
are  the  only  available  facts  from  which  to  produce  the 
required  bracket. 

In  Fig.  :;s-J,  let  M  X  or  <)  I'  be  drawn  at  the  re- 
quired angle,  with  reference  to  any  horizontal  line,  to 
represent  the  pitch  of  the  gable  cornice.  The  first  step 
is  to  redraw  the  normal  side  elevation  of  the  level 
bracket  so  that  its  vertical  lines  shall  be  at  right  angles 
to  the  lines  of  the  rake,  all  as  shown  at  L  Q  P.  Next, 
at  any  convenient  distance  from  this  draw  two  rertical 
lines,  as  M  ( )  and  N'  1",  the  horizontal  distance  between 
which  shall  be  the  required  face  width  of  the  bracket. 
Lines  projected  parallel  to  the  rake  from  the  various  j 
angles  in  the  profile  between  these  vertical  lines  will 
complete  the  front  elevation  of  the  raking  bracket.  The 
additional  lines  E  <!  and  F  II  representing  the  sink  in 
the  face,  A  C  showing  the  depth  of  the  panel  in  side, 
and  U  1)  giving  the  depth  of  the  sink  in  the  face,  will 
be  understood  from,  the  drawing. 

To  construct  a  side  view  of  the  raking  bracket,  or, 
what  is  the  same  thing,  the  pattern  for  the  side  (includ- 
ing the  bottom  of  the  sunken  panel  and  the  sink  strips 
I'  I)  Z  in  the  face),  all  hights  must  be  measured  upon 
one  of  the  vertical  lines  of  the  face  view,  as  M  0.  To 
avoid  confusion,  however,  and  make  room  for  other 
patterns,  another  vertical  line.  X'  P%  will  serve  as  well. 
Divide  the  curved  portions  U  to  P  of  the  face  of  the 
normal  profile  into  any  convenient  number  of  small 
spaces  for  use  in  this  and  subsequent  parts  of  the  oper- 
ation. From  all  the  points  in  the  profile  of  face  carry 
lines  '  parallel  to  the  rake  through  the  side  view  and 
continue  them  till  they  intersect  the  vertical  line  X1  P3. 
From  the  points  thus  obtained  in  the  line  X1  P3  carry 
lines  indefinitely  horizontally,  as  indicated.  Upon 
each  oi  the  lines  so  drawn  lay  off  from  the  line  X'  Ps  a 
distance  or  distances  equal  to  the  distance  or  distances 
upon  the  corresponding  lines  drawn  across  the  normal 
side  of  the  bracket.  Through  the  points  thus  obtained 
trace  lines,  which  will  give  the  several  shapes  in  the 
sides  of  the  brackets  corresponding  to  the  shapes  shown 


in  the  side  of  the  normal  bracket.  It  may  be  necessary 
to  introduce  in  the  several  profiles  of  the  normal  bracket 
other  points  than  those  derived  from  spacing  the  profile. 
Use  as  many  such  points  as  may  be  necessary  to  deter- 
mine the  position  of  all  points  in  the  side  being  con- 
structed. Then  X'  N"  P'  will  be  the  pattern  of  the  side 
of  the  bracket,  and  U"  Z"  D3  will  be  the  pattern  of  the 
strip  forming  the  sides  of  the  sink  shown  in  the  face  by 
E  F  II  G,  and  &'  a1  dl  c'  will  be  the  shape  of  the  panel  in 
the  side  of  the  bracket. 

For  the  patterns  of  the  several  pieces  forming  the 
face  of  the  bracket  the  profiles  are  to  be  found  in  the 
normal  side  view,  from  which  stretchouts  can  be  ob- 
tained when  wanted,  and  laid  out  at  right  angles  to  the 
lines  of  the  rake;  while  the  miter  lines  of  any  part  are 
the  vertical  lines  of  the  face  view  corresponding  to  that 
part  of  the  profile  under  consideration. 

For  the  strip  REGS,  forming  that  part  of  the 
face  at  the  side  of  the  sink,  lay  off  a  stretchout  of  its 
profile  U  Z  at  right  angles  to  the  lines  of  the  rake,  as 
shown  by  u1  z\  through  the  points  in  which  the  usual 
measuring  lines  are  drawn.  Drop  the  points  from  the 
profile  to  the  miter  lines  R  S  and  E  G;  then,  with  the 
T-square  placed  at  right  angles  to  the  lines  of  the  rake, 
and  brought  successively  against  the  points  in  R  S  and 
E  G,  the  corresponding  measuring  lines  are  cut.  Then 
lines  traced  through  these  points  of  intersection,  as  . 
shown  by  R1  S1  and  E1  G3,  form  the  pattern  for  that 
piece. 

For  the  piece  forming  the  face  of  the  bracket  below 
the  sink,  as  shown  in  the  elevation  by  S  0  P1  Z1,  pro- 
ceed in  like  manner.  A  stretchout  of  its  profile,  as 
indicated  by  D  P,  is  laid  off  at  right  angles  to  the  lines 
of  the  rake,  through  which  the  usual  measuring  lines 
are  drawn.  The  points  in  D  P  are  then  carried  parallel 
to  the  rake,  cutting  the  miter  lines  S  0  and  Z1  P'. 
The  T-square  is  then  placed  at  right  angles  to  the  lines 
of  the  rake,  and  brought  against  the  several  points  in 
the  sides  S  0  and  Z'  P',  by  which  the  corresponding 
measuring  lines  are  cut.  In  like  manner  it  is  brought 
against  the  points  G  and  II,  by  which  the  shape  of  the 
part  extending  up  to  meet  the  sink  is  determined. 
Then  lines  traced  through  these  several  points  of  inter- 
section, as  shown  by  IIs  Z3  Ps  0'  S3  G*,  form  the  pattern 
for  that  part  of  the  face  of  the  bracket.  The  upper 
part  of  the  face  of  the  bracket,  shown  in  the  face  view 


194 


The  New  Metal    HW.vr    l'nll>',-i<    Tiook, 


by  N1  U1  K  M,  being  u  Hat  surface,  as  indicated  in  the 
side  view  N  U,  is  obtained  by  pricking  directly  from 
the  face  view  of  the  bracket,  no  development  of  it 
being  necessary. 

To  avoid  confusion  of  lines,  the  sink  piece  E  FH 


prolilc,  as  shown  by  «'  d\  is  laid  oil  at  right  angles  to 
the  lines  of  the  rake,  and  through  the  points  in  it  the 
usual  measuring  lines  are  drawn.  The  "  -square  is  then 
placed  at  right  angles  to  the  lines  <>f  the  rake,  and,  be- 
ing brought  successively  against  the  points  in  the  sides 


«VP« 

\^.\\\\$^\\Y<L 


Fiij.  383.— Upper  Return  of 
Bracket  Head. 


Fig.  384.— Lower  Return  of 
Bracket  Head. 


Fig.  382.— The  Patterns  for  a  Raking  Bracket. 


G  is  transferred  to  the  right,  as  shown  by  E1  F1  H'  G'. 
The  profile  of  it,  as  indicated  in  the  side  view  by  UD, 
is  divided  into  any  convenient  number  of  spans. 


E1  G'  and  F'  II',  the  corresponding  measuring  lines  are 
cut.  Then  lines  traced  through  these  points  of  inter- 
section, as  shown  by  E3  G"  F2  IF,  constitute  the  pattern 


and    through    the  points  lines   are   drawn,  cutting  the      of  the  bottom  of  the  sink. 


miter  lines  E1  G'  and  F'  II'.     The   stretchout  of    this 


Of  the  strips  bounding  the  panel  of  the  side  in  the 


Pattern   Problems. 


195 


bracket,  ilie  piece  corresponding  to  l>  c  in  the  side 
view,  being  vertical,  is  obtained  by  pricking  directly 
from  its  elevation  in  llie  face  view  of  the  bracket,  A  B 
D'  C  being  the  shape.  For  the  other  straight  strip 
bounding  this  panel,  shown  in  the  side  view  by  a  b, 
the  length  is  laid  off  equal  to  a  b,  while  the  width  is 
taken  from  the  face  view,  equal  to  the  space  indicated 
by  A  B.  For  the  strip  representing  the  irregular  part 
a  to  c  proceed  as  follows :  Divide  the  profile  a  d  c  into 
any  convenient  number  of  parts,  from  the  points  in 
which  carry  lines  crossing  the  face  view  of  the  same 
part,  as  indicated  by  A  B  1)'  C.  At  right  angles  to 
the  lines  of  the  rake  lay  off  a  stretchout  of  the  profile 
just  named,  as  indicated  by  a'  c',  through  the  points 
in  which  draw  the  usual  measuring  lines.  Place  the 
1 -square  at  right  angles  to  the  lines  of  rake,  and, 
bringing  it  against  the  several  points  in  the  line  A  C 
and  B  D',  cut  the  corresponding  measuring  lines  drawn 
through  the  stretchout.  Then  lines  traced  through 
the  several  points  of  intersection  thus  formed,  as  indi- 
cated by  A'  G"  and  B1  D',  will  be  the  pattern  of  the 
curved  strip  forming  part  of  the  boundary  of  the  panel 
in  the  side  view  of  the  bracket. 

Of  the  three  pieces  of  molding  forming  the  head 
of  the  bracket,  the  profile  of  the  piece  across  the  face 
is  normal,  as  shown  at  L  N,  while  that  of  the  two  side 
pieces,  or  returns,  requires  to  be  modified  or  raked 
before  a  square  miter  with  the  face  piece  can  be 
eH'ec.tcd.  These  principles  will  be  further  explained 
in  Problems  91  to  94.  The  first  step  will  be  to  draw  a 
correct  elevation  of  the  head,  which  includes  raking  the 
profiles  of  the  upper  and  lower  returns. 

Divide  the  normal  profile  L  N  into  convenient 
spaces,  and  from  the  points  thus  obtained  carry  lines 
indefinitely  parallel  to  the  rake  across  the  top  of  the 
I'.K  e  view  of  the  bracket.  Draw  duplicates  of  the  nor- 
mal profile,  placing  them  in  a  vertical  position  directly 
above  where  the  new  sides  are  required  to  be,  as  shown 
b\-  it  I  and  k  m.  Divide  these  two  profiles  into  the 
same  number  of  parts  employed  in  dividing  the  normal 
profile,  and  from  these  points  drop  lines  vertically, 
intersecting  those  drawn  from  L  N.  Then  lines  traced 
through  these  points  of  intersection,  as  shown  by 
L'  N1  and  K  M,  will  be  respectively  the  profiles  of  the 
moldings  on  the  upper  side  and  on  the  lower  side  of  the 
bracket.  Lay  off  a  stretchout  of  the  profile  L  N  at 
right  angles  to  the  line  of  the  rake  and  through  the 
points  in  it  draw  the  usual  measuring  lines.  "With  the 
blade  of  the  T-square  at  right  angles  to  the  lines  of 
the  rake,  and  brought  successively  against  the  several 


points  in  the  profile  N'  L'  and  K  M,  cut  the  measuring 
lines  drawn  through  the  stretchout.  Then  lines  traced 
through  the  points  of  intersection  thus  obtained,  as 
shown  by  L3  N'  and  K1  M',  will  be  the  shape  of  the 
ends  of  the  molding  forming  the  front  of  the  bracket 
head. 

Before  laying  out  the  pattern  for  the  return  mold- 
ing forming  the  upper  side  of  the  bracket  head  a  cor- 
rect side  elevation  of  it  must  be  drawn.  A  duplicate 
of  the  profile  L'  N1  is  transferred  to  any  convenient 
place,  as  shown  at  L3  1ST*  in  Fig.  383,  and  parallel  lines 
from  its  angles  are  extended  to  the  right,  as  shown, 
making  L'  Q1  equal  to  L  Q  of  the  side  view  of  the 
bracket. 

At  Q1  repeat  the  outline  L"  N4,  which  represents 
the  intersection  of  the  bracket  head  with  the  bed  mold 
of  the  cornice.  L'  N1  of  Fig.  382  is  then  the  cor- 
rect profile  and  the  lines  L'  N4  and  Q1  X"  are  the  miter 
lines  of  this  return ;  however,  as  both  the  miter  lines 
are  identical  with  the  profile,  the  stretchout  q  x  may 
be  taken  from  either  one,  the  other  being  divided  into 
the  same  number  of  spaces  as  the  first,  which  is  easier 
than  dropping  the  points  from  one  to  the  other.  The 
T-square  may  then  be  placed  at  right  angles  to  the 
lines  in  the  molding  and  brought  successively  against 
the  points  in  the  lines  L3  N4  and  Q'  X*,  and  the  cor- 
responding measuring  lines  intersected.  Then  lines 
traced  through  these  points,  as  shown  .by  L4  N*  and 
Q3  X3,  will  form  the  pattern. 

The  pattern  for  the  return  molding  of  the  head 
occurring  on  the  lower  side  of  the  bracket  is  obtained 
in  the  same  manner.  A  duplicate  of  the  profile  K  M 
of  the  face  view  of  the  bracket  is  drawn  at  any  con- 
venient place,  as  shown  by  K3  M2  in  Fig.  384.  The 
proper  length  is  given  to  the  molding  by  measuring 
upon  the  side  view  of  the  bracket,  and  a  duplicate  of 
the  profile  is  drawn  at  the  opposite  end.  Space  the 
profile  K"  W  into  any  convenient  number  of  parts,  as 
indicated  by  the  small  figures,  and  in  like  manner  di- 
vide the  profile  K3  M3  into  the  same  number  of  parts. 
At  right  angles  to  the  line  of  the  molding  lay  off  a 
stretchout  of  these  profiles,  as  shown  by  k'  m1,  through 
which  draw  the  usual  measuring  lines.  With  the  blade 
of  the  T-srluarc  at  right  angles  to  the  lines  of  the 
molding,  and  brought  successively  against  the  several 
points  in  the  profiles  K3  M3  and  K3  M3,  cut  the  corre- 
sponding measuring  lines.  Then  a  line  traced  through 
these  points  of  intersection,  as  shown  by  K5  M5  and 
K4  M4,  will  constitute  the  pattern  of  the  return  mold- 
ing, or  the  lower  side  of  the  bracket. 


196 


Tlie  New  Metal    Worker  Pattern  Book. 


PROBLEM    89. 


The  Pattern  for  a  Raised  Panel  on  the  Face  of  a  Raking  Bracket. 


In  the  solution  of  the  problem  stated  above,  and 
which  is  given  in  Fig.  385,  the  first  requisite  is  tin- 
design  or  outline  of  the  side  of  the  normal  bracket,  as 
such  an  outline  is  really  a  section  through  the  raking 
bracket  upon  a  line  at  right  angles  to  the  rake.  N  S  T 
shows  the  side  view  of  a  normal  bracket,  or  the  bracket 
as  it  would  appear  in  a  level  cornice,  the  part  from  G 
to  H  being  molded  as  shown  by  the  shaded  profile, 
which  profile,  being  a  section  on  line  a  b  of  the  normal 
bracket,  is  given  complete  at  J  and  called  the  nor- 
mal profile.  The  first  step  is  to  derive  from  these 
factors  a  front  elevation  of  the  molded  panel  upon  tin- 
face  of  the  raking  bracket.  To  accomplish  this  first 
divide  the  profile  of  the  panel  molding  into  any  con- 
venient number  of  equal  parts,  as  shown  in  the  section 
shaded  in  the  side  of  the  normal  bracket,  and  through 
these  points  draw  lines  parallel  to  the  face  of  the 
bracket,  producing  them  iintil  they  cut  the  upper  sur- 
face against  which  the  panel  terminates,  and  in  the 
opposite  direction  until  they  meet  the  vertical  surface 
in  the  lower  part  of  the  bracket  against  which  the  panel 
terminates  at  the  bottom.  From  the  points  thus  ob- 
tained in  the  horizontal  surface  near  the  top  of  the 
bracket  and  in  the  vertical  surface  near  the  bottom  of 
the  bracket  draw  lines  at  right  angles  to  the  face,  thus 
transferring  the  points  to  the  line  representing  the  outer 
face  of  the  panel,  as  shown  from  G  to  II. 

These  points  will  be  used  a  little  later  in  develop- 
ing the  view  of  the  panel  at  rjght  angles  to  the  face. 
Next,  from  the  points  already  obtained  in  the  line  rep- 
resenting the  vertical  surface  near  the  bottom  of  the 
bracket  carry  lines  parallel  with  the  rake,  extending 
them  across  the  front  elevation  of  the  bracket.  In  the 
diagram,  to  avoid  confusion,  these  lines  terminate  at 
the  intersections  shown  from  A  to  B,  but  in  actual 
work  they  would  be  extended  across  the  front  eleva- 
tion, thereby  making  also  the  intersections  shown  from 
C  to  D.  At  any  convenient  place  in  line  with  the 
front  elevation  of  the  raking  bracket  draw  the  normal 
profile,  as  shown  below  the  elevation,  and  divide  it 
into  spaces  corresponding  to  the  spaces  used  in  dividing 
the  profile  in  the  side  view.  From  the  points  thus 
obtained  carry  lines  vertically,  intersecting  those  just 
drawn  from  the  side  of  the  normal  bracket  across  the 
front  elevation.  A  line  traced  through  the  points  of 


intersection  gives  the  outlines  shown  at  A  B  and  C  D. 
These  outlines  constitute  a  front,  elevation  of  the  lower 
end  of  the  molded  panel,  or  the  view  as  seen  from  u 
point  exactly  in  front  of  the  face  <>f  the  raking  bracket 
when  finished  and  in  its  proper  or  linal  position.  The 
outline  or  shape  of  the  upper  end  of  the  panel  \vouhl 
appear  as  a  simple  straight  line  in  this  view  because  it 
miters  against  a  surface  which  is  horizontal  from  front 
to  back.  ABC  I)  F  K  shows  the  entire  front  view  of 
the  molded  panel.  This  view  furnishes  the  means  for 
the  next  step,  which  is  to  obtain  a  view  at  right  angles 
to  the  face  G  H,  and  at  the  same  at  right  angles  to  the 
lines  of  the  rake  N  0.  To  do  this,  first  continue  the 
lines  from  the  normal  profile  of  panel  in  their  vertical 
course  till  they  intersect  the  upper  line  of  the  panel 
E  F.  These  lines  are  omitted  through  the  face  of  the 
bracket,  the  points  only  being  indicated  on  the  line  K 
F.  From  the  points  thus  established  in  K  K.  and  from 
the  points  derived  in  the  outlines  A  B  and  C  D,  carry 
lines  at  right  angles  to  the  raking  cornice,  producing 
them  indefinitely,  as  shown.  At  right  angles  to  the 
raking  cornice,  at  any  convenient  place,  draw  the  line 
H'  and  G1,  setting  off  on  it  spaces  corresponding  to 
those  established  in  II  G,  already  described.  Through 
the  points  in  H1  G1  draw  lines  at  right  angles  to  it  to 
the  left,  producing  them  until  they  intersect  lines  al- 
ivadv  drawn  from  the  outlines  A  B  and  C  D  and  the 
points  in  the  line  E  F.  Through  the  points  of  inter- 
section thus  obtained,  a.s  indicated  by  1  T  in  the  lower 
left  hand  corner,  8  14  in  the  lower  right  hand  corner,  s, 
9,  10,  etc.,  in  the  upper  right  hand  corner,  and  1,  2,  3, 
4,  etc.,  in  the  upper  left  hand  corner,  trace  lines,  thus 
completing  a  view  of  the  panel  piece  at  right  angles  to 
its  face.  The  next  step  to  be  taken  is  to  develop  a 
true  profile  of  this  panel,  or  in  other  words,  a  section 
at  right  angles  to  its  lines,  from  which  to  obtain  a 
stretchout  for  the  required  pattern.  To  do  this,  first 
assume  any  line,  as  P  0,  at  right  angles  to  the  lines  of 
the  view  just  obtained  as  the  surface  of  the  panel  in 
the  new  profile.  Upon  this  line  extended,,  as  at  K, 
draw  a  duplicate  of  normal  profile  so  that  the  points  7 
and  S  shall  lie  in  it.  Divide  the  profile  K  into  the 
same  number  of  spaces  as  in  previous  instances,  and 
from  these  points  carrv  lines  through  the  face  view  in- 
tersecting them  with  lines  of  corresponding  number,  as 


Pattern  Problems. 


197 


14 


NORMAL  PROFILE,       a 


Fig.  385.— The  Pattern  for  a  Raised  Panel  on  the  Face  of  a  Raking  Bracket. 


198 


Tlic  New  Metal    Worker  Pattern  Book. 


shown  at  L  P  and  Q  R.  Then  L  P  Q  K  will  be  the 
true  profile  of  the  moldings  along  the  face  of  the  raking 
bracket.  The  student  will  observe  that  only  half  the 
profile  is  shown  at  K,  as  both  halves  are  alike,  one- 
half  will  answer  all  purposes  if  it  be  kept  in  mind  while 
making  the  intersections  by  number  that  the  points 
1-7  in  one  profile  are  14-8  in  the  other.  At  any  con- 
venient place  lay  off  the  stretchout  of  the  true  profile, 


as  shown  to  the  left  by  the  line  L  M.  Through  the 
points  in  this  line  draw  the  usual  measuring  lines,  as 
shown.  Then,  with  the  blade  of  the  T-square  placed 
parallel  with  the  stretchout  line  and  brought  against 
the  several  points  of  intersection  at  the  corners  of  the 
"  View  at  Right  Angles  to  the  Face,"  cut  correspond- 
ing measuring  lines.  Lines  traced  through  the  points 
thus  obtained  will  produce  the  pattern  shape,  as  shown. 


PROBLEM   90. 

The  Patterns  for  a  Diagonal  Bracket  Under  Cornice  of  a  Hipped  Roof. 


In  Fig.  386  is  shown  a  constructive  section  of  the 
cornice  of  a  hipped  roof,  under  which  the  bracket  L 
fits  against  the  planceer  and  over  the  bed  molding  C. 
Fig.  387  shows  an  inverted  plan  of  the  angle  of  such  a 
cornice,  including  two  normal  brackets  B  and  C,  and 
the  diagonal  bracket  D,  of  which  the  patterns  are  re- 
quired. At  A,  in  line  with  one  arm  of  the  cornice  in 
plan,  is  also  shown  a  duplicate  of  the  profile  of  the 
normal  bracket.  E  F  represents  the  miter  line  of  the 
planceer  over  which  the  diagonal  bracket  is  required 
to  fit. 

Two  distinct  operations  are  necessary  in  obtaining 
the  patterns  of  the  bracket  D,  one  for  the  face  pieces 
and  the  other  for  the  sides.  As  the  bracket  is  placed 
exactly  over  (or  more  properly  speaking  under)  the 
miter  in  the  cornice,  one-half  its  width  must  be  drawn 
on  either  side  of  the  miter  line,  as  shown  in  Fig.  387. 
Each  half  of  its  face  thus  becomes  a  continuation  of 
the  moldings  forming  the  faces  of  the  course  of  normal 
brackets  of  which  it  is  a  part.  Therefore  the  normal 
profile  X  8  of  the  bracket  A  is  the  profile  to  be  used, 
and  I  G  and  J  F  form  the  miter  lines  for  one  half 
the  face. 

The  usual  method  in  obtaining  the  pattern  for  the 
face  piece  would  be  to  divide  the  profile  of  A  into  any 
convenient  number  of  spaces  and  lay  off  a  stretchout 
of  the  same  upon  any  line  drawn  at  right  angles  to  the 
direction  of  the  mold — that  is,  at  right  angles  to  I  Jor 
G  F — after  which  lines  should  be  dropped  from  the 
profile  upon  the  rniter  lines  and  thence  into-the  stretch- 
out. However,  as  the  miter  is  a  square  miter,  the 
short  method  is  available ;  hence  the  stretchout  line  is 
drawn  at  right  angles  to  the  horizontal  line  of  the  ele- 
vation X  X,  as  shown  at  II  G.  The  usual  measuring 
lines  are  drawn  and  intersected  with  lines  from  points 
of  corresponding  number  on  the  profile.  Lines  traced 


through  the  points  of  intersection,  as  shown  by  K  M 
and  L  N",  will  give  the  pattern  for  half  the  face. 

The  operation  of  obtaining  or  "  raking  "  the  pat- 
tern of  the  side  is  exactly  similar  to  that  employed  in 
Problem  88,  with  the  difference  that  while  in  Problem 
88  the  side  is  elongated  vertically,  in  the  present  in- 
stance (the  cornice  remaining  horizontal,  and  the  bracket 
being  placed  obliquely)  it  is  elongated  laterally  or 
horizontally.  The  operation  is  also  complicated  by 
the  addition  of  a  profile  at  the  back  edge  of  the  bracket 


Fig.  S80. — Sectional  View  of  the  Cornice  of  <t  Hipped  Hoof, 
Showing  Bracket, 

where  it  is  required  to  fit  over  the  bed  molding  of  the 
cornice.  To  obtain  the  pattern  of  the  side  it  is  first 
necessary  to  ascertain  the  correct  horizontal  distances 
between  the  various  points  of  the  profile.  The  points 
already  made  use  of  in  obtaining  the  face  may  be  used 
for  this  purpose.  Therefore,  drop  lines  from  each  of 
these  points  vertically,  intersecting  the  side  of  the 
bracket,  or,  what  is  the  same  thing,  the  center  line  E  F, 
as  shown  in  the  plan,  Fig.  387,  by  1',  2',  3',  etc.  The 


Pattern  Problems. 


profile  at  the  back  of  the  bracket  in  the  elevation  must 
als<>  be  divided  into  a  convenient  number  of  spaces,  as 
shown  by  the  small  figures,  which  must  also  be  dropped 
upon  E  F,  as  shown,  and  numbered  correspondingly. 


H 

-r 


transfer  the  points  and  spaces  from  E  F.  Low  from 
each  point  in  the  line  E'  F',  erect  lines  vertically,  in- 
tersecting lines  of  corresponding  number  previously 
drawn  to  the  right  from  the  elevation.  Thus,  lines 
drawn  upward  from  the  intersections  1,  2,  3,  4,  etc., 
on  the  line  E'  F'  intersect  with  horizontal  lines  1,  2,  3, 
4,  etc.,  while  lines  drawn  upward  from  the  intersec- 
tions 1',  2',  3',  4',  etc.,  on  the  line  E' F' intersect  with 
horizontal  lines  1',  2',  3',  4',  etc.  Lines  traced  through 
the  points  of  intersection,  as  shown  by  0  R  S  P,  will 
be  the  required  pattern  of  the  side. 

If  it  be  desirable  to  ascertain  the  exact  angle  to 
which  to  bend  the  edges  or  flanges  of  the  bracket  to 
fit  against  the  planceer  it  may  be  accomplished  in  the 
following  manner :  Extend  0  P  of  the  pattern  of  the 
side  till  it  intersects  the  line  from  X  of  the  side  eleva- 


Fig.  S87.— Inverted  Flan  of  Cornice  and  Method  of  Obtaining  Patterns. 


From  each  of  the  points  in  the  profile  of  the  elevation 
carry  lines  indefinitely  to  the  right,  as  shown.  At  any 
convenient  point  at  the  right  of  the  plan,  draw  another 
plan  of  the  diagonal  bracket,  so  placed  that  its  sides 
shall  be  parallel  with  the  horizontal  line  X  X  of  the 
elevation,  all  as  shown,  and  upon  its  center  line  E'  F' 


tion,  as  shown  at  Y.  Upon  the  solid  line  X  X  in 
diagonal  elevation  establish  any  point,  as  T.  Through 
the  point  T  and  at  right  angles  to  Y  P  draw  a  line  in- 
tersecting the  line  Y  P  at  U. 

As  the  angle  of  the  plan  shown  in  Fig.   387  is  a 
right  angle,  construct  a  right  angle,  ABC,  Fig.  388, 


200 


Tlie  New  Metal    Worker  Pattern  Book. 


and  bisect  it,  obtaining  the  miter  line  B  L.  Now  take 
the  distance  from  Y  to  T  in  diagonal  elevation,  and 
place  it  'on  the  miter  line  B  L  in  Fig.  388,  from  B  to 
D.  At  right  angles  to  B  L  draw  a  line  through  the 
point  D,  intersecting  the  sides  of  the  right  angle  A  B 


Fig.  S88.— Diagram  for  Obtaining  Angles  for  Sending  the  Flanges. 

C  at  E  and  F.  Now  take  the  distance  T  U  in  diag- 
onal elevation  and  set  it  off  from  D  toward  B,  locating 
the  point  H.  Connect  the  points  E,  II  and  F;  then 
will  the  profile  E  H  F  in  Fig.  388  represent  a  section 
across  the  hip  at  right  angle  to  its  rake  and  will  also 
be  the  angle  to  be  used  in  putting  the  straight  parts  of 


the  face  together,  as  shown  by  E'  H'  F'  in  Fig.  389. 
Tin-  anisic  which  the  sides  of  the  bracket  make  with 
the  planceer  will  be  the  complement  of  the  angle  II  K 
D  of  Fig.  388  and  may  be  obtained  as  follows:  Paral- 
lel to  B  L,  in  Fig.  388,  and  through  the  point  E,  draw 


Fig.  S89. — Perspective  View  of  Finished  Bracket. 

I  K,  representing  the  vertical  side  of  the  bracket ;  then 
will  the  angle  J  E  I  represent  the  profile  required  IW 
bending  the  flanges  on  the  side  of  the  bracket,  the  pro- 
file being  shown  in  position  by  I'  E'  J'  in  Fig.  389. 

In  Fig.  389  is  shown  a  perspective  view  of  the 
finished  bracket  as  seen  from  below. 


PROBLEM  91. 

To  Obtain  the  Profile  of  a  Horizontal  Return,  at  the  Foot  of  a  Gable,  Necessary  to  Miter  at  Right 
Angles  in  Plan  With  an  Inclined  Molding  of  Normal  Profile,  and  the  Miter  Patterns  of  Both. 


In  the  elevation  B  C  E  D,  and  plan  F  G  H  K  L  I, 
of  Fig.  390  is  presented  a  set  of  conditions  which 
necessitate  a  change  of  profile  in  either  the  horizontal 
or  raking  molding,  in  order  to  accomplish  a  miter 
joint  at  I  II  in  the  plan.  In  other  words,  the  condi- 
tions are  such  that  with  a  given  profile,  as  shown  by 
A1,  in  the  raking  molding,  the  profile  of  the  horizontal 
molding  forming  the  return  will  require  to  be  modi- 
fied, as  shown  by  the  profile  A*,  in  order  to  form  a 
miter  upon  the  line  I  H  in  the  plan. 

The  reason  for  this  is  easily  found.  If  a  vertical 
line  be  erected  from  point  9  in  profile  A°  it  will  be 
seen  that  each  line  emanating  from  a  point  in  the  nor- 
mal profile  A1  becomes  depressed  after  passing  this 
vertical  line,  more  or  less,  according  as  its  distance 
away  from  this  line  increases,  all  in  proportion  to  the 
amount  of  rake  or  incline  of  the  face  molding,  as 
shown  by  the  dotted  lines.  If,  on  the  contrary,  the 
profile  A"  be  considered  as  the  normal  profile,  the  pro- 


file A'  will  have  to  be  changed  or  "raked,"  in  this 
case  increased  in  hight,  in  proportion  to  the  inclina- 
tion. (These  conditions  are  treated  in  the  succeeding 
problem.)  The  vertical  hight  of  the  profile  of  the  re- 
turn may  be  measured  in  the  side  elevation  and  com- 
pared with  that  of  the  inclined  molding  by  measuring 
across  the  latter  at  right  angle's  to  the  line  B  C. 

In  this  problem  it  is  assumed  that  the  profile  as 
well  as  the  pitch,  or  rake,  of  the  cornice  B  C  are 
established  and  that  the  profile  of  the  horizontal  re- 
turn is  to  be  modified,  or  "raked,"  to  suit  it.  To 
obtain  this  profile,  first  draw  the  normal  profile  in  the 
raking  cornice,  as  shown  by  A1,  placing  it  to  corre- 
spond to  the  lines  of  the  cornice,  as  shown.  Draw 
another  profile  corresponding  to  it  in  all  parts,  directly 
above  or  below  the  foot  of  the  raking  cornice,  in  line 
with  the  face  of  the  new  profile  to  be  constructed, 
placing  this  profile  A  so  that  its  vertical  lines  shall  cor- 
respond with  the  vertical  lines  of  the  horizontal  cor- 


Pattern   Problems. 


201 


nice.  Divide  the  profiles  A  and  A'  into  the  same 
number  of  parts,  and  through  the  points  thus  obtained 
draw  lines,  those  from  A1  being 
parallel  to  the  lines  of  the  raking 
cornice,  and  those  from  A  intersecting 
them  vertically.  Through  these  points 
of  intersection  of  like  numbers  trace  a 
line,  which  gives  the  modified  profile, 
as  shown  by  A2.  Then  A3  is  the 
profile  of  the  horizontal  return,  indi- 
cated byG  II  1  F  in  the  plan.  It  is  also 
the  elevation  of  the  miter  line  I II  of  the 
plan.  Therefore  at  any  convenient  point 
at  right  angles  to  the  lines  of  the  raking 
cornice  lay  off  the  stretchout  M  N  of 
the  profile  A1,  through  the  points  in 
which  draw  measuring  lines  in  the 
usual  manner.  Place  the  J-^quare  at 
right  angles  to  the  lines  of  the 
raking  cornice,  and,  bringing  it  suc- 
cessivelv  against  the  points  in  the 
profile  A2,  cut  the  corresponding  measuring  lines 
just  described.  Through  the  points  of  intersection 
trace  a  line,  as  shown  by  O  P  II.  Then  OPE 
will  be  the  shape  of  the  lower  end  of  the  raking 
cornice  mitering  against  the  return.  For  the  pattern 
of  the  return  proceed  as  follows:  Construct  a  side 
elevation  of  the  return,  as  shown  by  S  V  U  T,  mak- 
ing the  profile  V  U  the  same  as  the  profile  A'  of  the 
elevation.  Let  the  length  of  the  return  correspond  to 
the  return,  as  shown  in  the  plan  by  F  I.  In  the  pro- 
file V.U  set  off  points  corresponding  to  the  points  in 
the  profile  A2  as  shown  from  B  to  D.  At  right  angles 
to  the  elevation  of  the  return  lay  off  a  stretchout  of 
V  U,  or,  what  is  the  same,  of  the  profile^A.2,  as  shown 
by  W  X,  through  the  points  in  which  draw  measuring 
lines  in  the  usual  manner.  Placing  the  T-square 
parallel  to  this  stretchout  line,  and  bringing  it  success- 


ively  against  the  points  in  V  U,  cut  tne  corresponding 
measuring  lines.      Then    a    line  traced   through,  these 

Z 


1 

Fig.  390.— To  OotzLi  tte  Pr.-f.le  cf  c,  Il.-rizontzl  return  at  the 
Foot  of  a  Gable,  Necessary  to  Miter  at  night  Angles  in  Plan  with 
an  Inclined  Molding  of  Normal  Profile,  and  the  Patterns  of  Both. 

points  of  intersection,  as  usual,  from  Y  to  Z,  will  be 
•the  pattern  of  the  horizontal  return. 


PROBLEM  92. 

To  Obtain  the  Profile  of  an  Inclined  Molding  Necessary  to  Miter  at  Right  Angles  in  Plan  with  a  Given 

Horizontal  Return,  and  the  Miter  Patterns  of  Both. 


The  conditions  shown  in  this  problem  are  similar 
to  those  in  the  one  just  demonstrated.  In  this,  how- 
ever, the  normal  profile  is  given  to  the  horizontal  re- 
turn, and  the  profile  or  the  raking  cornice  is  modified 


to  correspond  with  it.  To  obtain  the  new  profile  pro- 
ceed as  follows:  Divide  the  normal  profile  A1,  Fig. 
391,  into  any  convenient  number  of  parts  in  the  usual 
manner,  and  from  these  points  carry  lines  parallel  to 


202 


Tlte  New  Metal    Worker  Pattern  Book. 


the  lines  of  the  raking  cornice  indefinitely.     At  any 
convenient  point  outside  of  the  raking  cornice,  and  at 


F' 


4 

/ 

\ 

| 

<? 

/ 

7 

ID 

9 

l! 

M 

H'    Sidi 

> 

1 
K 

1 

( 

• 

* 

1 

2 

3 

4 

[/ 

5 

F 

/ 

7 
8 

• 

— 

- 

7 

9 

10 

/ 

ii 

12 

/ 

13 

-I2 

4 

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Plan 


Fig.  391.— To  Obtain  the  Profile  nf  an  Inclined  Mold- 
ing Necessary  to  Miter  at  Rir/ht  Angles  in  Plan 
with  a  Given  Horizontal  Return,  and  the  Pat- 
terns of  Both. 


right  angles  to  its  lines,  construct  a  duplicate  of  the 
normal  profile,  as  shown  by  A",  which  divide  into  like 


number  of  spaces.      With  the  T-square  at  right  angles 
to  the  lines  of  the  raking  cornice,  and  brought  suen  sa- 
ively   against    the  several    points    in    this  profile,    cut 
corresponding  lines  drawn  through  the  cornice  from 
the  profile  A1.       Then   a   line    traced    through    these 
points  of  intersection,   as  shown  by.  A",  will  be  the 
profile  of  the  raking  cornice.     For  the  pattern  of  the 
foot  of  the  raking  cornice  mitering  against  the  return, 
take  the   stretchout  of  the  profile  A3  and  lay  it  off  on 
any  line  at  right  angles  to  the  raking  cornice,  as  shown 
by  P  0.      Through  the  points  in  this  stretchout  line 
draw  the  usual  measuring  lines,  as  shown.     With  the 
T-square  at  right  angles  to  the  lines  of  the  raking  cor- 
nice, or  parallel  to  the  stretchout  line,  bring 
it  successively  against  the  points  in  the  pro- 
file A',    which    is    also  an  elevation  of  the 
miter,    and    cut  the  measuring  lines  drawn 
through  the  stretchout   P   O.     Then  a  line 
traced  through  the  points  of  intersection,  as 
shown  by  B'  R1,  will  be    the  miter  pattern 
of  the  foot  of  the  raking  cornice. 

For  the  pattern  of  the  return  proceed 
as  follows :  Construct  an  elevation  of  the 
return,  as  shown  by  F'  G'  K1  H',  in  di- 
mensions making  it  correspond  to  F  G  K  II 
of  the  plan.  Space  the  profile  A  of  the  re- 
turn in  the  same  manner  as  A'.  At  right  angles  to 
the  lines  in  the  return  cornice  draw  any  straight  line, 
as  M  N,  on.  which  lay  off  its  stretchout,  through  the 
.  points  in  which  draw  measuring  lines  in  the  usual 
manner.  Place  the  T-square  at  right  angles  to  the 
lines  of  the  return  cornice,  and,  bringing  it  successively 
against  the  points  in  the  profile  A,  cut  the  correspond- 
ing measuring  lines.  Through  the  points  of  intersec- 
tion ti'ace  a  line,  as  shown  by  G2  K".  In  like  manner 
draw  a  line  corresponding  to  F1  II1  of  the  side  eleva- 
tion. Then  F2  G"  Ks  IP  will  be  the  pattern  of  the 
horizontal  return  to  miter  with  the  raking  cornice,  as 
described. 


PROBLEM   93. 

To  Obtain  the  Profile  of  the  Horizontal  Return  at  the  Top  of  a  Broken  Pediment  Necessary  to  Miter 
with  a  Given  Inclined  Molding:,  and  the  Patterns  of  Both. 


In  Fig.  392,  C  B  D  represents  a  portion  of  the 
elevation  of  what  is  known  as  a  "broken  pediment," 
the  normal  profile  of  whose  cornice  is  shown  at  A1. 
With  these  conditions  existing  it  becomes  necessary  lo 


obtain  new  profiles  for  the  returns  at  both  the  top  And 
the  foot.  The  method  of  raking  the  return  at  the  foot 
has  been  described  in  Problem  91,  and  the  method  of 
raking  the  return  at  the  top  is  exactly  the  same.  If, 


Pattern  Problems.  . 


203 


in  the  designing  of  the  pediment,  the  normal  profile 
should  be  placed  in  the  return  at  the  foot,  as  is  some- 
times necessary,  then  the  profile  <>f  tin-  inclined  mold- 
ing must  be  first  obtained,  which  in  turn  must  be  con- 
sidered as  a  normal  profile  and  used  as  a  basis  of  ob- 
taining the  third  profile,  that  of  the  return  at  the  top. 

In  Fig.  392,  let  A'  be   considered   as   the  normal 
profile  of  the  inclined  molding.      Divide  A1  into- any 
convenient  number  of  parts  in  the  usual  manner,  and 
through  these  points  draw  lines  parallel  to  the  lines  of 
the  cornice  indefinitely.-    At  any  convenient  point  out- 
side of  the  cornice,  and  in  a  vertical  line  with  the  point 
at  which  the  new  profile  is  to  be  constructed,   draw  a 
duplicate  of  the  profile  of  the  raking  cornice,  as  shown 
l»v  A,  which  space  into  the  same   number  of  parts  as 
A',  already  described.      From  the  points  in  A   draw 
lines  vertically,  intersecting  lines  drawn  from  A1.   Then 
a  line  traced  through  these  several  points  of  intersec- 
tion, as  shown  by  A3,  will  constitute  the  profile  of  the 
horizontal  return  at  the  top  and  also  the  miter  line  as 
shown  in  elevation.      If  the  normal  profile  were  in  the 
horizontal   return  at  the  foot  of  the  pediment  and  the 
modified  profile  in  the  position  of  A1, 
it    would    he    immaterial    whether  the 
normal    profile    or    a  duplicate  of    the 
modified  profile  were  in  the    place   of 
A  by  which  to  obtain   the  intersecting 
lines,    as   the  projection    of  the  points 
only  is  to  be  considered  in  this  opera- 
tion, and  that  is  the  same  in  both  cases. 

For  the  pattern  of  the  inclined 
molding  proceed  as  follows  :  At  right 
angles  to  the  lines  of  the  raking 
cornice  lay  off  a  stretchout  of  the  profile  of  the  raking 
cornice  A1,  as  shown  by  F  G,  through  the  points  in 
which  draw  measuring  lines  in  the  usual  manner. 
Place  the  T-square  at  right  angles  to  the  lines  of  the 
raking  cornice,  and,  bringing  the  blade  successively 
against  the  points  in  the  profile  A",  which  is  the  miter 
line  in  the  elevation,  cut  the  corresponding  measuring 
lines,  and  through  these  points  of  intersection  trace  a 
line,  as.  shown  by  G  II.  Then  G  II  will  be  the  pat- 
tern of  the  top  end  of  the  raking  cornice  to  miter 
against  the  horixontal  return.  For  the  pattern  of  the' 
horizontal  return  the  usual  method  would  be  to  con- 
struct an  elevation  of  it  in  a  manner  similar  to  that 
described  for  the  return  at  the  foot  of  the  gable  in  the 
preceding  demonstrations  ;  the  equivalent  of  this,  how- 
ever, can  be  done  in  a  way  to  save  a  considerable  por- 
tion of  the  labor. 


As  the  view  of  the  miter  line  is  the  same  in  both 
the  front  and  the  side  elevation  the  pattern  may  be  de- 
veloped from  the  front  just  obtained  in  the  following 
manner,  with  the  result,  however,  that  the  pattern  will 
be  reversed :  Draw  the  line  K  M  perpendicular  to  the 


Fig.  SOS.— To  Obtain  the  Profile  of  the  Horizontal  Return  •  '  the 
Top  of  a  Broken  Pediment  Necessary  to  Miter  with  a  Given 
Inclined  Molding,  and  the  Patterns  of  Both. 


lines  of  the  horizontal  return,  as  it  would  be  if  shown 
in  elevation.  Upon  K  M  lay  off  a  stretchout  of  the 
profile  A1,  all  as  shown  by  the  small  figures,  and 
through  the  points  draw  the  usual  measuring  lines. 
With  the  T-square  parallel  to  the  stretchout  line  K  M 


204 


Tlie  New  Metal   Worker  Pattern  Book. 


bring  the  blade  successively  against  the  points  in  the 
profile  A%  cutting  the  corresponding  measuring  lines. 
Through  these  points  of  intersection  trace  a  line,  as 


shown  by  N  L,  which  will  be  the  pattern  of  the  end 
of  the  horizontal  return  to  miter  against  the  trable 
cornice,  as  shown. 


PROBLEM  94. 

To  Obtain  the  Profile  and  Patterns  of  the  Returns  at  the  Top  and  Foot  of  a  Segmental  Broken  Pediment. 


The  preceding  three  problems  treat  of  the  various 
miters  involved  in  the  construction  of  angular  pedi- 
ments. In  Fig.  393  is  shown  an  elevation  of  a  curved 
or  segmental  broken  pediment  in  which  the  normal 
profile  is  placed  in  the  horizontal  return  at  the  foot. 
The  profiles  for  the  curved  molding  and  for  the  return 
at  the  top  can  both  be  obtained  at  one  operation  in 
the  following  manner :  Divide  the  normal  profile 
ABC  into  any  convenient  number  of  parts,  and  from 
the  points  thus  obtained  draw  lines  at  right  angles  to 
the  horizontal  line  C  F  of  elevation,  as  shown.  At 
any  convenient  point  draw  G  II,  at  right  angles  to 
A  Gr,  cutting  them.  With  Q,  the  point  from  which 
the  curve  of  the  molding  was  struck,  as  center,  strike 
arcs  from  the  points  in  A  B  C,  extending  them  in  the 
direction  of  D  indefinitely.  From  any  convenient 
point  in  the  arc  A  D,  as  L,  draw  a  line  to  the  center  Q. 
From  L  draw  L  M,  at  right  angles  to  L  Q,  upon  which, 
beginning  at  L,  set  off  the  distances  contained  in  II  G, 
as  shown  by  the  small  figures  in  L  M.  From  the 
points  of  intersection  where  arcs  struck  from  Q  cut 
L  Q  draw  lines  at  right  angles  to  L  Q.  From  the 
points  in  L  M,  and  at  right  angles  to  it,  drop  lines 
cutting  those  of  similar  number  drawn  at  right  angles 
to  L  Q.  A  line  traced  through  these  points  of  inter- 
section, as  shown  by  M  K,  will  be  the  profile  of 
curved  molding.  It  will  be  observed  that  the  points 
for  obtaining  the  profile  are  where  he  perpendiculars 
dropped  from  L  M  intersect  the  lines  drawn  at  right 
angles  to  L  Q,  and  not  where  the  perpendiculars 
dropped  from  L  M  intersect  the  arcs. 

For  the  profile  D  E  draw  N  D,  parallel  to  C  J,  or 
at  right  angles  to  N  0,  and,  starting  from  D,  set  off 
on  D  N  the  same  points  as  are  in  G  II.  Drop  perpen- 
diculars from  these  points  to  the  arcs  of  similar  num- 
bers drawn  from  A  B,  when  a  line  traced  through  the 
points  of  intersection  will  form  the  desired  profile,  as 
show  by  D  E.  The  normal  profile  is  also  drawn  above 
G  II  and  N  D  at  X  and  Z  to  show  that  the  same  result 
is  obtained  by  using  the  points  in  G  II  to  set  off  on 
L  M  and  N  D  as  would  be  obtained  by  dropping  the 
points  from  the  profiles.  The  patterns  for  the  returns 


would  be  obtained  as  described  in  the  previous  prob- 
lems. 


Fig.  393.— To  Obtain  the  Profiles  and  Patterns  of  the 
Returns  at  the  Top  and  Foot  of  a  Segmented  Broken 
Pediment. 


Problems  describing  the  method  of  obtaining  the 
pattern  for  the  blank  for  the  curved  molding  will  be 
found  in  Section  2  of  this  chapter. 


Pattern  Problems. 


205 


PROBLEM  95. 

From  the  Profile  of  a  Given  Horizontal  Molding,  to  Obtain  the  Profile  of  an  Inclined  Molding  Necessary 
to  Miter  with  it  at  an  Octagon  Angle  in  Plan,  and  the  Patterns  for  Both  Arms  of  the  Miter. 


Another  example  wherein  is  required  a  change  of 
prolilc  in  order  to  produce  a  miter  between  the  parts 
is  shown  in  Fig.  394.  In  this  case  the  angle  shown 


us  indicated,  and  in  the  corresponding  side,  as  shown  in 
elevation  by  N  OLK,  draw  a  duplicate  profile,  as  shown 


by  A'. 


S94.—  From  the  Profile  of  a  Given  Horizontal 
Moldiiiy  to  Obtain  the  Profile  of  an  Inclined 
Moldiny  Necessary  to  Miter  n-ith  it  at  an  Octa- 
gon Angle  in  Plan,  and  the  Patterns  fur  Roth 
Arms  of  the  Miter. 


iii  plan  between  the  abutting  members  is  that  of  an 
octagon,  as  indicated  by  BCD.  To  produce  the 
modified,  profile  and  to  describe  the  patterns  proceed 
as  follows  :  In  the  side  B  0  draw  the  normal  profile  A, 


Divide  both  of  these  profiles  into  the  same  num- 
ber of  parts,  and  from  the  points  in  each 
carry  lines  parallel  to  the  lines  of  mold- 
ing in  the  respective  views,  producing 
the  lines  drawn  from  profile  A  until 
they  meet  the  miter  line  C  X.  From 
the  points  thus  obtained  in  C  X  erect 
lines  vertically  until  they  meet  those 
drawn  from  profile  A',  intersecting  as 
shown  from  0  to  L.  Through  these 
points  of  intersection  draw  the  line 
O  L,  which  will  be  the  miter  line  in 
elevation  corresponding  to  C  X  of  the 
plan.  From  the  points  in  O  L  carry 
lines  parallel  with-  the  raking  molding 
in  the  direction  of  P  indefinitely.  At 
any  convenient  point  outside  of  the 
raking  cornice  draw  a  duplicate  of  the 
normal  profile,  as  shown  by  A2,  placing 
its  vertical  line  at  right  angles  to  the 
lines  of  the  raking  cornice.  Divide  the 
profile  A2  into  the  same  number  of 
spaces  as  employed  in  A  and  A1,  and 
from  these  points  carry  lines  at  right 
angles  to  the  lines  of  the  raking  cornice, 
intersecting  those  of  corresponding 
number  drawn  from  the  points  in  0  L. 
Trace  a  line  through  these  intersections, 
as  shown  from  E  to  S.  Then  R  S 
will  be  the  required  profile  of  a  raking 
cornice  to  miter  against  a  level  cornice 
of  the  profile  A  at  an  angle  indicated  by 
B  C  D  in  the  plan,  or  an  octagon  angle. 
For  the  pattern  of  the  level  cornice, 
at  right  angles  to  the  arm  B  C  in  the 
plan  lay  off  a  stretchout  of  the  profile 
A,  as  shown  by  E  F,  through  the 
points  in  which  draw  the  usual  measur- 
ing line.  With  the  T-square  at  right 
angles  to  B  C,  bringing .  the  blade  suc- 
cessively against  the  several  points  in 
X  C,  cut  corresponding  measuring  lines  drawn  through 
I1!  !•'.  Then  a  line  traced  through  these  points,  as  shown 
from  II  to  G,  will  be  the  required  pattern  of  the  hori- 
zontal cornice.  In  like  manner,  for  the  pattern  of 


206 


The  Xew  Metal    Worker  Pattern  Book. 


the  raking  cornice,  at  right  angles  to  its  lines  lay  off  a 
stretchout  of  the  profile  R  S,  as  shown  l>y  U  T, 
through  the  points  in  which  draw  measuring  lines  in 
the  usual  manner.  With  the  T-square  at  right  angles 
to  the  lines  of  the  raking  cornice,  arid  brought  success- 


ively against  the  points  in  the  miter  line  O  L,  as  shown 
in  elevation,  cut  the  corresponding  measuring  lines. 
Then  a  line  traced  through  the  points  thus  obtained, 
us  shown  by  W  V,  will  be  the  required  pattern  for  the 
raking  cornice. 


PROBLEM   96. 

From  the  Profile  of  a  Given  Inclined  Molding,  to  Establish  the  Profile  of  a  Horizontal  Molding  to  Miter 
with  it  at  an  Octagon  Angle  ia  Plan,  and  the  Patterns  for  Both  Arms. 


In  Fig.  395,  let  B  C  D  be  the  angle  in   plan   at 
which  the  two  moldings  are  to  join,  U  0   V  the   angle 
in  elevation,   and  A  or  A1   the  nor- 
mal   profile    of    the    raking    mold. 
To  form  a  miter  between  moldings 
meeting    under   these    conditions    a 


VL 


Fig.  SOS.— From  the  Profile  of  a  Given  Iiu-lined  Moldiwj. 
to  Establish  the  Profile  of  a  Horizontal  Moldimj 
to  Miter  with  it  at,  an  Octagon  Angle  in  Plan,  and 
the  Patterns  for  Both  Arms. 


change  of  profile  is  required.      To  obtain  the  modified 
profile  for  the  horizontal  arm  and  the  miter  line  in 


elevation  proceed  as  follows:  Draw 
the  normal  profile  A  with  its  vertical 
side  parallel  to  the  lines  in  the  plan  of 
the  arm  E  X  D  C,  corresponding  to  the 
front  of  the  elevation.  Draw  a  dupli- 
cate of  the  normal  profile  in  correct 
position  in  the  elevation,  as  shown 
by  A1.  Divide  both  of  these  profiles 
into  the  same  number  of  parts,  and 
through  the  points  in  each  draw  lines 
parallel  with  the  plan  and  with  the 
elevation  respectively,  all  as  indicate.] 
by  the  dotted  lines.  From  the  points 
in  the  miter  line  of  the  plan  C  E, 
obtained  by  the  lines  drawn  from  the 
profile  A,  carry  lines  vertically,  inter- 
secting the  lines  drawn  from  A1. 
Then  a  line  traced  through  the  inter- 
sections thus  obtained,  as  shown  from 
N  to  0,  will  be  the  miter  line  in  ele- 
vation. From  the  points  in  N  O 
carry  lines  horizontally  along  the  arm 
of  the  horizontal  molding  N  O  U  Y, 
as  shown.  At  any  convenient  point 
outside  of  this  arm,  either  above  or 
below  it,  draw  a  duplicate  of  the  nor- 
mal profile,  as  shown  by  A",  which 
divide  into  the  same  number  of  parts 
as  before,  and  from  the  points  carry 
lines  vertically  intersecting  the  lines 
drawn  from  N  0,  just  described 
Then  a  line  traced  through  tin •.-(• 
points  of  intersection,  as  shown  by 
T  S,  will  give  the  required  modified 
profile. 

For  the  patterns  of  the  arm  Y  N 
O  U  proceed  as  follows:  At  right 
angles  to  the  same,  as  shown  in  plan 
by  W  E  C  B,  lay  off  on  any  straight  line,  as  G  F.  a 
stretchout  of  the  profile  T  S,  all  as  sh'own  by  the  small 


Pattern  Problems. 


207 


figures  I3,  2",  3",  etc.  Through  these  points  draw 
measuring  lines  in  the  usual  manner.  With  the  T-square 
parallel  to  the  stretchout  line,  and  brought  against  the 
points  of  the  miter  line  E  C  in  plan,  cut  corresponding 
measurirg  lines,  as  indicated  by  the  dotted  lines,  and 
through  these  points  of  intersection  trace  a  line,  as 
shown  by  K  II.  Then  K  II  will  be  the  shape  of  the 
end  of  Y  N  0  U  to  miter  against  the  raking  molding. 

It  will  be  easily  understood  that  the  points  as 
found  iipon  the  line  E  C  arc  just  the  same  as  would  be 
obtained  there  if  the  newly  obtained  profile  were  drawn 
into  the  plan  of  the  arm  C  13  W  E  and  the  points  were 


dropped  from  it  to  the  line  E  C  according  to  the  rule. 
For  the  pattern  of  the  raking  molding,  at  right  angles 
to  the  arm  N  Z  V  O  in  the  elevation  lay  out  a  stretch- 
out, L  M,  from  the  profile  A'.  Through  the  points 
in  this  stretchout  draw  measuring  lines  in  the  usual 
manner.  Place  the  T-square  parallel  to  the  stretchout 
line,  and,  bringing  it  against  the  several  points  in  the 
miter  line  in  elevation  N  0,  cut  corresponding  measur- 
ing lines,  as  indicated  by  the  dotted  lines.  Then  a 
line  traced  through  these  points  of  intersection,  as 
shown  by  P  R,  will  be  the  shape  of  the  cut  on  the  arm 
N  Z  V  0  to  miter  against  the  horizontal  molding. 


PROBLEM  97. 

The  Miter  Between  the  Moldings  of  Adjacent  Gables  of  Different  Pitches  upon  a  Pinnacle  with  Rectangular  Shaft. 


The  problem  presented  in  Figs.  396  and  397  is 
one  occasionally  arising  in  pinnacle  work.  The  figures 
represent  the  side  and  end  elevations  of  a  pinnacle  which 


Firj.  396.— Side  Elevation  of  Rectangular  Pinnacle,  Showing  the 
Miter  Between  the  Moldinijs  of  Adjacent  Gables. 

is   rectangular,  but    not    square.      All  of  its  faces  are 
tiuished  with  gables  whose  moldings  miter  with  each 


other  at  the  corners,  and  which  are  of  the  same  hight 
in  the  line  of  their  ridges,  as  indicated  by  L  M  and  L' 
M1.  Whatever  profile  is  given  to  the  molding  in  one 
face  of  such  a  structure,  the  profile  of  the  gable  in  the 
adjacent  face  will  require  some  modification  in  order  to 
form  a  miter.  In  Fig.  396  let  A  be  the  normal  profile 
of  the  molding  placed  in  the  gable  of  the  side  elevation. 
Before  the  miter  patterns  can  be  developed  it  will  first 
be  necessary  to  obtain  the  miter  line  or  joint  between 
the  moldings  of  the  adjacent  gables  as  it  will  appear 
in  the  elevation,  to  accomplish  which  proceed  as  fol- 
lows :  Draw  a  duplicate  of  A,  placing  it  in  a  vertical 
position  directly  below  or  above  the  point  at  which  the 
two  moldings  are  to  meet,  as  shown  by  A1.  Divide 
both  of  these  profiles  into  the  same  number  of  parts,  as 
indicated  by  the  small  figures,  and  through  these  points 
draw  lines  intersecting  in  the  points  from  H  to  K,  as 
shown.  Then  a  line  traced  through  these  intersections 
will  be  the  miter  line  in  elevation.  For  the  pattern  of 
the  molding  of  the  side  gable  lay  off  at  right  angles  to 
II  M  a  stretchout  of  the  profile  A,  as  shown  by  B  C, 
through  the  points  of  which  draw  the  usual  measuring 
lines.  Place  the  T-square  at  right  angles  to  the  lines 
of  the  molding,  or,  what  is  the  same,  parallel  to  the 
stretchout  line,  and,  bringing  it  against  the  several 
points  in  the  miter  line  II  K,  cut  corresponding  meas- 
uring lines.  Then  a  line  traced  through  these  points,  as 
shown  by  D  E,  will  be  the  shape  of  the  cut  at  the  foot 
of  the  side  gable  to  miter  against  the  adjacent  gable. 

The  next  step  is  to  obtain  the  correct  profile 
of  the  molding  on  the  adjacent  gable.  H  K  having 
been  established  as  the  correct  elevation  of  the  miter,  its 


208 


The  Xcw  Mdal    Wurkvr  Pattern  Bovk. 


outline  may  now  be  transferred,  with  its  points,  to  the 
end  elevation  of  the  pinnacle,  as  shown  tit  II1  K',  Fig. 
397,  reversing  it,  because  it  appears  here  at  the  right 
side  of  the  gable,  whereas  it  appeared  at  the  left  of  the 
other.  Draw  a  duplicate  of  the  normal  profile,  as  shown 
at  A",  placing  its  vertical  lines  at  right  angles  to  the 
lines  of  the  gable,  and  divide  it  into  the  same  spaces  as 
in  the  first  operation.  From  these  points  draw  lines  at 
right  angles  across  the  molding,  which  intersect  with 
lines  drawn  parallel  to  the  molding  from  the  points  in 
the  miter  line  IP  K'.  Then  a  line  traced  through 
these  points  of  intersection  will  form  the  required  modi- 
fied profile,  as  shown  by  W  X. 

For  the  pattern  of  the  molding  of  the  end  gable 
proceed  as  follows :  At  right  angles  to  the  lines  of  the. 
raking  cornice  lay  off  a  stretchout  of  the  profile  W  X, 
as  shown  by  P  R,  through  the  points  in  which  draw 
measuring  lines  in  the  usual  manner.  With  the 
T-square  at  right  angles  to  the  lines  of  the  raking 
cornice,  bringing  it  successively  against  the  points  in 
K1  H1,  cut  corresponding  measuring  lines.  Then  aline 
traced  through  these  points  of  intersection,  as  shown 
from  S  to  T,  will  be  the  pattern  required. 


Fly.  S97. — End  Elevation  of  Rectangular  Pinnacle,  Slwn-ing  Same 
Miter  as  in  Fig.  396. 


PROBLEM  98. 

The  Miter  Between  the  Moldings  of  Adjacent  Gables  01'  Different  Pitches  upon  an  Octagon   Pinnacle. 


This  problem  differs  from  the  preceding  one  in 
that  the  angle  of  the  plan  is  octagonal  instead  of 
square,  but  like  it  requires  a  change  of  profile  in  one 
of  the  gables  in  order  to  effect  a  miter.  In  Figs.  398 
and  399  are  shown  a  quarter  plan  of  pinnacle  and  the 
elevations  of  two  adjacent  gables  of  different  widths 
but  of  similar  bights.  Let  A1  B'  F'  O  G1  D'  of  Fig. 
398  be  a  correct  elevation  and  A  B  C  G  be  a  quarter 
plan  of  the  structure.  In  that  portion  of  the  plan  cor- 
responding to  the  part  of  the  elevation  shown  to  the 
front  draw  the  normal  profile  E,  placing  its  vertical 
side  parallel  to  the  lines  of  the  plan.  Divide  it  into 
any  convenient  number  of  spaces,  and  through  these 
points  draw  lines  parallel  to  the  lines  of  the  plan,  cut- 
ting C  O',  the  miter  line  in  plan,  as  shown.  In  like 
manner  plaee  a  duplicate  of  the  normal  profile,  as  shown 
by  E1  in  the  elevation.  Divide  it  into  the  same  num- 
ber of  equal  parts,  aud  through  the  points  draw  lines 


parallel  to  the  lines  of  the  raking  cornice,  which  pro- 
duce in  the  direction  of  N  O  indefinitely.  Bring 
the  T-square  against  the  points  in  C  O1,  and  with  it 
erect  vertical  lines,  cutting  the  lines  drawn  from  E1,  as 
shown  from  N  to  O.  Then  a  line,  NO,  traced  through 
these  points  of  intersection  will  be  the  miter  line  in 
elevation. 

For  the  pattern  of  the  miter  at  the  foot  of  the 
wide  gable  or  gable  shown  in  elevation  proceed  as  fol- 
lows:  At  right  angles  to  the  lines  of  the  gable  cornice 
lay  off  a  stretchout  of  the  profile  E1,  as  shown  by  II  K, 
through  the  points  in  which  draw  the  usual  measuring 
lines.  Placing  the  T-square  at  right  angles  to  the  lines 
of  the  cornice,  or,  what  is  the  same,  parallel  to  the 
.stretchout  line,  and  bringing  it  against  the  several 
points  in  N  O,  cut  corresponding  measuring  lines. 
Then  a  line  traced  through  the  points  of  intersect:. <:•. 
thus  obtained,  as  shown  from  L  to  M,  will  be  the  put- 


P/ltf/'/'ll     Frnlil,  in    . 


209 


tern  of  the  miter  at  the  foot  of  the  gable  shown   in  ele- 
vation.     Fur  the  mudilied  profile  of  the  gable  molding 


/    'HIS.— Quarter  Plan  nnd  Elevation  of  Octayon    Pinnacle, 
iiirj  Miter    Bctu-een    Moldinrjs   of  Adjacent    (tables  of  Different 
Pitches. 


upon  the  narrow  side  proceed  as  follows:  Draw  n  cor- 
rect elevation  of  the  narrow  side,  reproducing  therein 


the  miter  line  N  ()  from  Fig.  398  (reversing  tne  same), 
as  shown  by  K  P  in  Fig. 399,  and  through  the  points, 
also  reproduced  from  N  O,  carry  lines  parallel  to  the 
lines  of  the  gable  cornice  indefinitely,  as  shown.  Draw 
a  duplicate  of  the  normal  profile  at  any  convenient 
point  outside  of  the  gable  cornice,  as  shown  by  E', 
placing  its  vertical  side  at  right  angles  to  A"  11,  or  the 
lines  of  the  cornice.  Divide  E3  into  the  same  number 
of  parts  as  used  in  the  other  profiles,  and  through  the 
points  draw  lines  at  right  angles  to  the  lines  of  the  cor- 
nice, intersecting  the  lines  drawn  from  P  R.  Through 


Fig.  S'JO. — Elevation  of  Narrow  Side  of  Octagon  Pinnacle,   Showing 
Same  Miter  as  in  Fig.  398. 


these  points  trace  a  line,  as  indicated  by  E3,  which  will 
be  the  modified  profile. 

To  lay  out  the  pattern  take  the  stretchout  of 
E3  and  lay  it  off  on  any  straight  line  drawn  at  right 
angles  to  the  lines  of  the  cornice,  as  S  T,  and  through 
the  points  in  it  draw  the  usual  measuring  lines.  Place 
the  T-square  at  right  angles  to  the  lines  of  the  gable 
cornice,  and,  bringing  it  against  the  points  in  P  R,  cut 
the  measuring  lines,  as  indicated  by  the  dotted  lines. 
Then  a  line  traced  through  these  points  of  intersection, 
as  shown  by  U  T,  will  be  the  pattern  for  the  molding 
at  the  foot  of  the  gable  on  narrow  side. 


210 


Tlie  New  Metal    Worker  Pattern  Book. 


PROBLEM  99. 

The  Patterns  for  a  Cold  Air  Box  in  which  the  Inclined  Portion  Joins  the  Level  Portion  Obliquely  in  Plan. 


The  conditions  of  the  problem  are  clearly  shown 
in  the  plan  and  side  elevation  of  Fig.  400,  in  which  Z 
B  C  is  the  elevation  and  X  C'  D'  Y  is  the  plan  of  the 
level  portion  of  a  cold  air  passage  joining  a  furnace 
just  above  the  floor  line.  The  inclined  portion  of  the 
air  passage  or  box  is  required  to  join  the  level  portion 
at  the  angle  Z  A  E  of  the  side  elevation,  and  at  the 
angle  Y  A'  E'  when  viewed  in  plan.  These  conditions 
are  in  many  respects  similar  to  those  given  in  Problem 
95,  with  the  difference,  however,  that  in  this  case  the 
joint  or  miter  between  the  level  and  the  inclined  por- 
tions does  not  appear  as  a  straight  line  in  the  plan.  It 
may  be  here  remarked  that  the  solution  of  this  prob- 
lem is  more  a  matter  of  drawing  than  of  pattern  cutting, 
as  nothing  can  be  more  simple  than  the  cutting  of  a 
miter  between  two  pieces  of  rectangular  .pipe  when  the 
required  angle  between  them  is  known.  This  problem 
is  capable  of  two  solutions,  both  of  which  will  be 
given,  leaving  the  reader  to  choose  which  is  the  more 
adaptable  to  his  requirements. 

First  Solution. — As  above  intimated,  before  the 
pattern  can  be  developed  it  will  be  necessary  to  make 
tareful  drawings,  in  the  preparation  of  which  a  knowl- 
edge of  the  principles  of  orthographic  projection  is 
necessary.  (See  Chapter  III). 

To  proceed,  then,  with  the  drawings,  first  draw  a 
plan  and  elevation  of  as  much  of  the  furnace  as  is 
necessary  to  show  its  connection  with  the  cold  air  box, 
placing  each  part  of  the  plan  directly  under  its  corre- 
sponding part  in  the  elevation,  so  that  as  soon  as  any 
new  point  is  determined  in  either  of  the  views  its  posi- 
tion can  be  located  in  the  other  by  means  of  a  perpen- 
dicular line  dropped  from  one  view  to  the  other.  Upon 
the  plan  set  off  the  width  of  the  box  b  and  draw  parallel 
lines  from  the  side  of  the  furnace  body  to  the  right 
indefinitely,  and  upon  the  elevation  set  off  its  hight,  a, 
from  the  floor  line  up,  and  draw  A  Z.  A  vertical  line 
from  the  point  X  of  the  plan  will  give  the  point  Z  upon 
the  elevation,  or,  in  other  words,  show  how  far  the 
curve  of  the  furnace  body  cuts  into  the  top  and 
bottom  surfaces  of  the  cold  air  box.  Next,  upon  the 
elevation  locate  the  point  A  the  required  distance  from 
the  side  of  the  body  according  to  specification  and  find 
its  position  in  the  plan  by  means  of  a  vertical  line,  as 
shown.  From  the  point  A  in  both  views  lines  must 
be  drawn  to  represent  the  angle  or  deflection  of  the 


pipe  as  it  would  appear  in  those  views.  Thus  the  ele 
ration  would  show  the  slant,  which  is  determined  by 
the  two  dimensions  c  and  d.  Therefore  from  the  point 
A  of  the  elevation  erect  a  perpendicular  line  equal  to 
the  required  hight  c,  from  the  top  of  which  draw  a 
horizontal  line  to  the  right  of  a  length  equal  to  the 
amount  of  slant  </,  thus  locating  the  point  K.  which 
connect  by  a  straight  line  with  A.  Then  will  A  E 
represent  the  angle  of  the  inclined  portion  of  the  pipe 
as  it  appears  in  the  elevation.  But  according  to  the 
requirements  the  pipe  is  also  to  have  an  oll'set  a  dis- 
tance equal  to  e — that  is,  the  point  E  of  the  elevation 
is  nearer  the  observer  than  the  point  A.  Therefore 
from  A'  of  the  plan  draw  a  line  forward  the  amount  of 
the  offset,  from  the  end  of  which  draw  a  line  to  the 
right,  in  length  equal  to  d,  or  in  other  words  till  it 
comes  directly  under  the  point  E  of  the  elevation,  thus 
locating  that  point  in  the  plan,  and  draw  A  E  .  which 
will  show  the  apparent  angle  in  the  plan. 

The  depth  and  width  of  the  oblique  portion  of  the 
box  will  next  demand  attention.  At  right  angles  to 
the  line  A  E  of  the  elevation  set  off  the  depth  of  the 
box  a  and  draw  a  line  to  represent  the  lower  near  cor- 
ner of  the  box,  which  continue  downward  until  it  cuts 
the  floor  line,  as  shown  at  D;  then  draw  A  D,  which 
represents  the  miter  cut  for  the  side  of  the  box.  At 
right  angles  to  A'  E'  of  the  plan  set  off  the  width  /»,  as 
shown,  and  draw  a  line  parallel  to  A'  E'  intersecting  the 
line  from  X  at  B',  as  shown,  and  draw  A'  B',  which 
gives  the  plan  of  the  miter  cut  across  the  top  of  the 
box.  As  the  point  D  of  the  elevation  is  in  the  same 
vertical  plane  as  A  it  may  now  be  dropped  into  the 
plan,  intersecting  with  the  line  showing  the  front  side 
of  the  box  in  that  view,  as  shown  at  D';  and  the  point 
B'  of  the  plan,  being  on  a  level  with  A',  may  be  pro- 
jected into  the  elevation,  where  it  would  intersect  with 
the  line  showing  the  top  of  the  box  at  B.  A  line 
drawn  from  D'  of  the  plan  parallel  to  A'  K'  (shown 
dotted)  will  then  show  the  position  of -the  lower  near 
angle  of  the  inclined  portion  of  the  box,  and  a  line 
from  B  of  the  elevation  parallel  to  A  E  will  show  the 
position  in  that  view  of  the  further  top  corner  of  the 
box. 

The  position  in  the  two  views  of  the  remaining 
angle  of  the  inclined  portion  of  the  box  may  be  ascer- 
tained in  several  ways:  The  width  b  may  be  set  off 


I '<dli  rn    /'rMnns. 


211 


from  1)'  of  tlic  plan  ami  :i  line  drawn  which  will  inter- 
sect with  X  B'  continued,  as  shown  at  C';  thence  it 
may  be  projected  into  the  elevation  at  C,  as  shown ; 
or  the  width  a  may  be  set  off  from  B  of  the  elevation, 
thus  locating  the  line  which  intersects  with  the  floor  at 
0,  which  point  may  be  dropped  into  the  plan,  thus 
locating  the  point  C' ;  or,  again,  B  G  may  be  drawn 
parallel  to  A  D,  or  1)'  C'  may  be  drawn  parallel  to  A' 
B',  all  producing  the  same  result. 

In   the  case   in   Problem  95,  above  referred  to,  it 
was  noted  that  if  the  normal  profile  is  adhered  to  in  the 
level  arm,  the  profile  of  the  gable  mold  must  be  changed 
or  "  raked"   before  a  perfect  miter   joint  can  be  ob- 
tained.     \Vhat  is  Irno  in  the  case,  of  the  gable  miter  is 
equally  true  in  the  case  of  the  furnace  pipe — a  correct 
profile  or  cross  section  of  the  box  must  be  developed 
in  order  that  a  correct  stretchout  may  be  obtained  for 
use  in  cutting  the  miter  of  the  inclined  arm  of  the  pipe. 
As  neither  the  plan  n<>r  the  elevation,  which  have  been 
correctly  obtained,  gives  the  true  length  of  the  inclined 
piece —that  is,  flic  true  distance  from  A   to  E — it  will 
be  necessary  to  obtain  still  another  elevation,  in  which 
such   distance   is  correctly  shown.      As  A'  E'  of  the 
plan  gives  the  liori/.ontal  distance  between  the 
points    A  and  K,  and  c  represents    the   vertical 
distance  between  them,  if  a  right  angled  triangle 
lie    constructed  with   A' 
K'    as    a    base    and    the 
hight  c  as  the  perpendic- 
ular, itshypothenuse  will 
then     give     the     desired 
measurement.       Such     a 
triangle    properly    forms 
part  of  an  oblique  eleva- 
tion which  may  be  pro- 
jected from   the  plan  in 
the    following    manner: 
Parallel  to  A'  E',  at  any 
convenient     distance 
away,  draw  a  line  to  rep- 
resent  the  level    of   the 
floor,    as    shown ;   above 
which,  at  a  distance  equal 
to  «,  draw  another  paral- 
lel line,  X"  A2,  represent-  . 
ing  the  hight  of  the  hori- 
zontal arm   of   the  pipe. 
Above  the  line  X"  A3,  at 
a   hight  equal  to  c,  draw 
still    another    line,  upon 
which  the  point  E  is  sub- 
sequently to  be  located. 


Fig.  400.  —Plan  and  Elevations  of  a  Cold  Air  Box  iti  Which  the  Inclined  Pcrtion  Joins  the  Level 
Portion  Obliquely  in  Plan. — First  Solution, 


212 


Tlie  Xt-w  Metal    Worker  Pattern  Book, 


Now  drop  lines  from  all  the  points  of  the  plan  at  right 
angles  to  A'  E',  intersecting  each  with  its  correspond  ing 
line  of  the  new  elevation,  thus  locating  each  point  of  the 
miter  in  that  view  As  points  D'  and  C'  are  upon  the 
floor,  their  position  will  be  found  at  D3  and  C2.  Like- 
wise lines  from  A'  and  B'  will  locate  those  points  in 
the  upper  surface  of  the  horizontal  pipe,  as  shown  at 
A*  and  Ba,  where  they  are  also  shown  to  be  in  the  side 
elevation.  A  line  dropped  from  E'  will  also  locate  that 
point  at  its  proper  hight,  as  shown  at  Ea.  A  line  con- 
necting A"  and  E3  will  then  be  the  hypothenuse  above 
alluded  to  and  be  the  correct  length  sought.  As  all 
edges  or  corners  of  the  pipe  are  necessarily  parallel, 
lines  drawn  from  B2  C'  and  D'  parallel  to  A'  EJ  will 
complete  this  part  of  the  elevation  as  far  as  necessary. 
In  these,  as  in  all  geometrical  drawings,  lines  showing 
parts  concealed  from  view  by  other  parts  are  always 
shown  dotted.  Lines  from  X  and  Y  locate  those  points 
in  the  new  elevation  and  show  that,  while  a  correct 
elevation  of  the  inclined  arm  of  the  pipe  has  been  ob- 
tained, the  view  of  the  horizontal  portion  is  oblique,  the 
space  between  X"  and  Y"  showing  the  open  end  to  fit 
against  the  furnace  body. 

Having  now  obtained  a  correct  oblique  elevation, 
the  next  step  is  to  obtain  a  correct  profile  upon  any 
line,  as  F  H,  drawn  at  right  angles  across  the  pipe, 
which  may  be  accomplished  in  the  following  manner : 
From  each  point  upon  the  line  of  the  section  F,  G,  J 
and  H  project  lines  parallel  with  the  direction  of  the 
pipe  to  a  convenient  point  outside  the  elevation,  as 
shown  at  the  left,  across  which  draw  a  line,  x  y,  at 
right  angles  to  them  as  a  base  from  which  to  measure 
distances  from  front  to  back. 

Assuming  its  crossing  with  the  line  from  G  (point 
1)  to  represent  the  near  angle  of  the  pipe,  set  off  from 
x  on  the  line  from  F  the  horizontal  breadth  of  the  pipe 
6,  thus  locating  point  4,  which  corresponds  to  the 
point  F  in  the  elevation.  In  like  manner  on  the  line 
from  II  set  off  from  y  the  distance  o  of  the  plan,  locat- 
ing the  point  2,  which  corresponds  to  point  II  of  eleva- 
tion, and  draw  the  lines  1  4  and  1  2.  The  distance 
of  point  3  from  line  x  y  is  equal  to  distance  b  plus  the 
distance  o,  or  in  other  words,  draw  the  line  2  3  par- 
allel to  1  4  and  the  line  4  3  parallel  to  1  2,  thus 
locating  the  point  3. 

Having  now  a  profile  and  a  correct  elevation 
of  the  miter,  nothing  remains  but  to  lay  off  a 
stretchout,  as  shown,  upon  the  line  H  K  and  drop 
the  points  in  the  usual  manner  from  the  profile  to 
the  miter  line  A5  B"  C''  D",  thence  into  the  measuring 


lines    of    the  stretchout,  all  as  clearly  shown  in    the 
drawing. 

As  the  plan  shows  all  the  dimensions  of  the  hori- 
zontal arm  of  the  pipe,  the  pattern  for  that  can  be  de- 
veloped in  the  usual  manner.  To  avoid  confusion  a 
duplicate  of  that  part  of  the  plan  has  been  transferred 
to  Fig.  401,  where  a  stretchout  of  the  normal  profile 
is  laid  off  at  right  angles  to  the  lines  of  the  pipe,  into 
which  the  points  are  dropped  from  the  miter  line  A  1? 
C  D.  In  the  normal  profile  of  course  the  distances  1 
4  and  2  3  are  equal  to  I  and  the  distances  1  2  and  4 
3  equal  to  a  of  Fig.  400. 


*>       D  12 

PLAN 
Fig.  401.—  Plan  and  Pattern  of  Level  Arm  of  Cold  Air  Boar. 

It  may  be  noted  here  that,  as  is  the  case  in  all 
raked  profiles,  the  dimensions  and  shape  of  the  profile 
obtained  from  the  oblique  elevation  differ  somewhat 
from  .those  of  the  normal  profile  shown  in  Fig.  401, 
and  that  their  stretchouts  are  therefore  necessarily  dif- 
ferent. 

Second  Solution.— It  may  be  asked  naturally,  is 
there  no  way  of  producing  a  miter  without  a  change  of 
profile,  just  as  a  carpenter  would  saw  off  the  ends  of 
two  square  sticks  of  timber  of  the  same  section  and 
produce  a  perfect  miter  at  an  oblique  angle  ?  There 
is,  but  the  method  of  doing  it  is  not  so  apparent  as  the 


Pattern    Problems. 


213 


one  just  described.  To  accomplish  this  a  drawing  or 
view  must  be  obtained,  in  which  the  surface  of  the 
paper  represents  a  plane  commou  to  both  arms  of  the 


shown  in  Fig.  402,  in  which  the  plan  shown  in  Fig. 
400  has  been  reproduced,  but  turned  around  in  such  a 
manner  as  to  facilitate  the  projection  from  it  of  an  end 


Fig.  402.—Pfiiterns  of  Cold  Air  Box. —Second  Solution. 

pipe.  As  three  points  determine  the  position  of  a 
plane,  it  will  be  seen  at  once  that  such  a  plane  passes 
through  the  points  Z,  A  and  E  of  the  side  elevation, 
Fig.  400.  The  best  means  of  obtaining  this  view  is 


elevation,  all  of  which  is  clearly  shown  in  the  drawing. 
This  view  shows  the  offset  e  and  the  rise  c  of  the 
oblique  portion  of  the  pipe.  The  new  view,  which 
will  give  the  required  conditions,  is  obtained  by  look- 
ing at  the  pipe  in  a  direction  at  right  angles  to  A  E  of 
the  end  elevation,  and  is  obtained  as  follows :  Parallel 
to  A  E  at  any  convenient  distance  away  draw  A'  E', 
which  make  equal  to  A  E  by  means  of  the  lines  drawn 
at  right  angles  to  A  E,  as  shown.  Upon  the  line  E'  E 
set  off  from  E'  the  slant  d  as  given  in  the  side  eleva- 
tion and  plan,  Fig.  400,  locating  the  point  E",  and 
draw  the  line  E2  A'.  From  all  points  of  the  profile  or 
end  view  of  the  horizontal  pipe,  1,  2,  3  and  4,  project 
lines  also  at  right  angles  to  A  E,  continuing  them 
across  the  line  A'  E',  and  make  A'  Y*  equal  to  A  Y 
of  the  plan.  Then  A'  Y"  will  be  the  length  of  the 
horizontal  arm  in  the  new  view  and  A'  E"  will  be  the 
length  of  the  inclined  arm,  both  lying  in  the  same 
plane,  and  the  angle  E"  A'  Y"  will  be  the  angle  at  which 
the  two  arms  meet.  Under  the  above  conditions,  then, 
a  line  which  bisects  that  angle,  as  A'  C,  will  be  the 


t)ie  New  Metal   Worker  Pattern  Hook. 


miter  line  between  the  two  arms.  As  the  two  arms  of 
the  miter  are  symmetrical,  the  view  can  be  completed, 
if  desired,  by  drawing  lines  parallel  with  A'  E2  from 
the  points  of  intersection  with  the  lines  from  the  end 
view  with  the  miter  line  A'  C.  As  1234  is  the 

• 

profile  from  which  the  short  arm  was  projected  in  the 
new  view,  a  stretchout  may  now  be  taken  from  it  and 


laid  off  on  any  line  at  right  angles  to  C  W  and  the  points 
dropped  in  the  usual  manner,  all  as  shown.  If  desired. 
the  stretchout  may  also  be  laid  off  at  right  angles  to 
the  inclined  arm  and  the  pattern  for  this  piece  thus 
developed  from  the  same  miter  line,  although  the  miter 
cut  A  B  C  D  A  is  the  same  in  both  pieces,  one  simply 
being  the  reverse  of  the  other. 


PROBLEM  100. 
The  Patterns  for  the  Inclined  Portion  of  a  Cold  Air  Box  to  Meet  the  Horizontal  Portion  Obliquely  in  Plan. 


This  problem  is  here  introduced  on  account  of 
the  similarity  of  its  conditions  with  those  of  the  one 
immediately  preceding,  although,  as  its  patterns  are 
obtained  entirely  without  the  use  of  profiles,  it  does 
not  properly  belong  in  this  connection.  Its  solution 
will  serve  to  show  what  widely  different 
means  may  be  employed  to  obtain  the 
same  ends.  In  the  preceding  case  the 
miter  cut  was  obtained  without  refer- 
ence to  the  miter  at  the  upper  end  of 
the  oblique  arm.  In  this  case  the 
oblique  portion  is  required  to  join,  at 
its  upper  end,  with  another  arm  like 
and  exactly  parallel  with  the  arm  join  • 
ing  its  lower  termination. 

Under  such  conditions  it  follows 
that  the  planes  of  the  upper  and  lower 
miters  must  be  parallel,  and,  therefore, 
that  miter  cut  at  the  upper  end  of 
either  of  the  faces  of  the  oblique  por- 
tion must  be  parallel  with  that  at  the 
lower  end  of  the  same.  Advantage 
may  be  taken  of  these  conditions  to 
obtain  a  very  simple  solution  of  the 
problem,  as  will  be  seen  below. 

The  first  requisite  is,  of  course,  a 
correctly  drawn  elevation  and  plan  in 
which  all  the  points  in  each  are  duly 
projected  from  corresponding  points  in 
the  other  view.  In  Fig.  403  is  shown 
a  plan  and  elevation  of  the  box,  with 
the  lines  of  projection  connecting  cor- 
responding points  in  each,  all  of  which  may  be  con- 
structed very  much  as  described  in  the  preceding 
problem. '  The  inclined  arm  is  required  to  have  a  rise 
equal  to  a  of  the  elevation  and  a  forward  projection 
equal  to  I  of  the  plan.  Corresponding  points  in  the 


two  views  are  lettered  alike.  Thus  the  elevation  shows 
clearly  that  it  is  an  elevatio:i  of  the  front  A  B  F  E  of 
the  plan,  with  the  back  C  I)  II  G  dotted  behind,  while 
the  plan  shows  clearly  A  B  D  G  of  the  elevation  with 
the  bottom  E  F  H  Gr  clotted  below. 


Fig.  JOS.— Patterns  for  the,  Inclined  Portion  of  a   CoM,  Air  Pox  to  Meet  the  Level 
Portion  Obliquely  in  Plan. 


The  first  important  information  to  be  derived  from 
the  correctly  drawn  views  is  that  the  front  and  back 
are  the  same,  likewise  the  top  and  bottom  are  alike. 
The  patterns  of  the  top  and  front  are  given  separately, 
upon  the  supposition  that  joints  will  be  made  at  all  of 


Pattern  Problems, 


SIB 


the  angles ;  should  they  he  wanted  in  one  piece  they 
could  readily  be  connected.  As  all  the  surfaces  of  the 
inclined  portion  of  the  pipe  are  oblique  to  the  given 
view,  only  some  of  their  dimensions  will  be  correct  as 
they  appear  on  the  paper.  An  inspection  of  both 
elevation  and  plan  will  show  that  the  lines  A  C  and 
13  D  are  both  horizontal  and  parallel,  and,  therefore, 
correct  as  they  appear  in  the  plan,  and  may  be  used 
as  given  in  the  construction  of  a  pattern  of  the  top 
piece.  The  shortest  distance  between  these  two  lines 
will  be  represented  by  a  line  at  right  angles  to  both, 
as  M  N.  Since  the  point  N  in  the  line  B  D  is  higher 
than  the  point  M  of  the  lino  A  C,  by  the  distance  a 
of  the  elevation,  it  will  be  necessary  to  construct  the 
diagram  J  L  K  in  order  to  get  the  correct  distance  be- 
tween the  points  M  and  N.  J  K  is  made  equal  to  the 
distance  M  X,  as  indicated  by  the  dotted  lines.  K  L 
is  equal  to  the  rise  given  in  the  elevation;  hence  the 
distance  J  L  represents  the  true  distance  between  the 
points  M  and  N.  Upon  the  continuation  of  the  line 
M  X  of  the  plan  set  off  the  distance  J  L,  as  shown  at 
,1 '  L'.  Through  each  of  these  points  lines  are  drawn 
parallel  to  A  0  and  B  D  of  the  plan.  The  line  A'  C' 
is  made  equal  to  A  C,  and  B'  I)'  is  made  equal  to  B  D 


by  means  of  the  dotted  lines  drawn  parallel  to  M  N. 
This  pattern  is  completed  by  connecting  the  point  A' 
with  B'  and  C'  with  D'. 

In  developing  the  pattern  of  the  side  A  B  F  E 
the  same  course  might  be  pursued,  beginning  with  the 
lines  A  E  and  B  F,  whose  lengths  are  correctly  given 
in  the  elevation,  but  for  the  sake  of  diversity  another 
method  has  been  employed.  Beginning  with  the 
known  fact  that  the  point  B  is  higher  than  the 
point  A,  as  shown  by  a  in  the  elevation,  con- 
struct a  diagram,  0  P  K,  making  O  P  equal  to 
and  parallel  with  A  B  of  the  plan,  and  0  E  equal 
to  a,  thus  giving  K  P  as  the  correct  length  of  the 
line  represented  by  A  B  of  the  plan.  From  the 
points  E  and  F  draw,  at  right  angles  to  E  F,  the  lines 
E  S  and  F  T  indefinitely.  Since  the  distances  A  E 
and  B  F  are  the  same  and  are  correctly  given  in  the 
elevation,  take  that  distance  between  the  feet  of  the 
dividers,  and  placing  one  foot  at  the  point  E  describe  a 
small  arc,  cutting  the  line  E  S  in  the  point  S.  By 
repeating  this  operation  from  the  point  P,  the  point  T 
is  established  in  the  line  F  and  T.  Lines  connecting 
the  points  E  S,  S  T  and  T  P  will  complete  the  pat- 
tern of  the  front  and  back. 


PROBLEM   101. 

The  Pattern  of  a  Hip  Molding:  upon  a  Right  Angle  in  a  Mansard  Roof,  Mitering  Against  the  Planceer 

of  a  Deck  Cornice. 


Let  Z  X  Y  V  in  Fig.  404  be  the  elevation  of  a 
deck  cornice,  against  the  planceer  of  which  a  hip  mold- 
ing,  shown  in  elevation  by  U  W  Y  T,  is  required  to 
miter.  Let  the  angle  of  the  roof  be  a  right  angle,  as 
shown  by  the  plan  Q  I)  A',  Fig.  405,  D  N"  represent- 
ing the  plan  of  the  angle  over  which  the  hip  molding 
is  to  be  placed.  This  angle  is  also  shown  bv  B  A  of 
the  elevation.  As  the  only  view  which  will  show  the 
correct  angle  at  which  the  hip  molding  meets  the  plan- 
ceer is  a  view  at  right  angles  to  the  line  D  N,  the  first 
step  in  the  development  of  the  patterns  will  be  to  con- 
struct such  a  diagonal  elevation.  Assume  any  point, 
as  A,  in  the  elevation  on  any  line  representing  a  plain 
surface  in  the  profile  of  -the  roof,  as  B  A.  Through  A 
draw  a  horizontal  line  indefinitely,  as  shown  by  LAC. 


From  B,  the  point  in  which  the  line  A  ±5  meets  the 
planceer,  drop  a  vertical  line,  cutting  the  horizontal 
line  drawn  through  A  at  the  point  C,  all  as  shown  by 
B  C.  Produce  the  line  of  planceer  W  Y,  as  shown  by 
W  Y'.  Draw  a  duplicate  of  the  plan,  Q  D  A'  in  Fig. 
405,  in  such  a  manner  that  the  diagonal  line  DN  shall 
lie  parallel  to  the  horizontal  line  drawn  through  A,  all 
as  shown  by  Q1  D1  A'.  At  right  angles  to  the  line  D' 
A",  at  any  convenient  point,  as  A5,  draw  the  line  A" 
C',  in  length  equal  to  the  distance  A  C  in  elevation, 
and  through  C'  draw  a  line  parallel  to  D1  A",  as  shown 
by  I  N1,  cutting  the  diagonal  line  I)1  N1  in  the  point 
N1.  Then  D1  N'  represents  the  diagonal  plan  of  that 
part  of  the  hip  from  B  to  A  in  the  elevation.  From 
N1  erect  a  perpendicular,  N1  M,  which  produce  until  it 


216 


Tlie  New  Metal    Worker  Pattern  Book. 


meets  the  line  carried  horizontally  from  the  plancccr 
in  the  point  B'.  In  like  manner  from  I)1  erect  a  per- 
pendicular, which  produce  until  it  meets  the  horizontal 
line  L  C  in  the  point  L.  Connect  L  and  B',  as  shown, 
which  will  constitute  the  desired  oblique  projection 
of  A  B. 

The  next  step  will  be  to  construct  a  section  of  the 
hip  molding  u-pon  a  line  at  right  angles  with  it,  as  G 


the  side  D'  A1,  from  which  erect  a  line  perpendicular 
to  D1  N',  as  shown  by  K  K,  which  produce  until  it 
meets  the  horizontal  line  L  C  in  the  point  L1,  and 
thence  carry  it  upward  parallel  to  L  B1,  cutting  G  II 
in  the  point  F1.  On  either  side  of  F1  lay  off  a  space 
equal  to  F  E  of  the  diagonal  plan,  as  shown  by  F'  K' 
and  F1  E".  Through  these  points  E'  and  E"  draw  lines 
to  K,  the  intersection  of  the  lines  L  B'  and  G  II.  From 


Fig.  404.— The  Pattern  of  a  Hip  Molding  in  a  Mansard  Roof,  Mitering  Against,  the  Planceer  of  a  Deck  Cornice. 


II,  assumed  at  any  convenient  point.  It  might  be 
supposed  that  in  such  a  section  the  two  fascias  of  the 
hip  molding  would  be  at  right  angles  to  each  other,  as 
they  undoubtedly  would  appear  in  the  plan  Q  D  A'  of 
Fig.  405  or  in  a  section  on  any  horizontal  line,  as  L  M. 
The  object  of  this  part  of  the  demonstration  is  to  show 
exactly  what  that  angle  would  be  and  how  to  obtain  it. 
Assume  any  point  in  the  diagonal  plan,  as  K,  in 


K  as  a  center  describe  the  curve  of  the  roll  of  the  re- 
quired diameter.  Upon  the  lines  K  E'  and  K  E"  ccs 
off  from  K  a  distance  sufficient  to  make  the  desired 
width  of  fascia,  thus  completing  the  profile  of  the  hip 
molding  in  the  diagonal  elevation. 

Space  one-half  of  this  profile,  as  G  EJ,  in  the  usual 
manner,  through  the  points  in  which  carry  lines  par- 
allel to  L  B',  cutting  the  line  of  plam-ccr  W  V,  which 


Pnaern  Problems. 


217 


is  the  miter  line  of  the  roll.      The  edges  of  the   fascia 
will  of  course  miter  with  the  lower  edge  of  the  fascia 


Fig.  405.— Plan  of  the  Fascias  and  Angle  of  the  Mansard  Shown 
in  Fig.  404. 


at  the  top  of  the  mansard,  shown  in  profile  at  B  E3,  all 
as  shown  by  the  dotted  lines  projected  from  E3.     At 


right  angles  to  the  line  L  B'  draw  the  straight  line  S  R, 
upon  which  lay  off  a  stretchout  of  the  profile  in  the 
usual  manner,  and  through  the  points  draw  measuring 
lines.  With  the  T-square  parallel  to  this  stretchout 
line,  or,  what  is  the  same,  at  right  angles  to  the  lines 
of  the  molding  in  the  diagonal  elevation,  and,  bringing 
it  successively  against  the  points  in  W1  Y1,  cut  corre- 
sponding measuring  lines  drawn  through  the  stretchout. 
The  measuring  lines  7  and  8  are  cut  from  the  inter- 
section of  the  fascia  of  the  hip  with  lines  projected 
from  E3  as  above  explained.  Then  a  line  traced 
through  these  points,  as  shown  in  the  engraving, 
as  shown  by  J1  P  0  J,  will  be  the  pattern  of 
the  hip  molding  inhering  against  the  horizontal 
planceer. 


PROBLEM  102. 

The  Pattern  for  a  Hip  Molding  upon  a  Right  Angle  in  a  Mansard  Roof,  Mitering  Against  a  Bed  Molding 

at  the  Top. 


Let  A  C  B,  in  Fig.  406,  be  the  section  of  a  por- 
tion of  a  mansard  roof,  the  elevation  of  which  is  shown 
to  the  left,  and  let  P  E  be  any  bed  molding  whose 
profile  does  not  correspond  to  or  member  with  the 
molding  used  to  cover  the  hips,  a  section  of  the  hip 
molding  being  shown  at  Z  Y  C'. 

The  solution  of  this  problem  will  be  accomplished 
by  means  of  a  "true  face"  of  the  roof,  rather  than 
by  means  of  a  diagonal  elevation  as  in  the  problem 
immediately  preceding  this.  Therefore,  supposing 
the  section  A  C  B  to  give  the  correct  pitch  of  the 
roof,  the  first  step  will  be  to  obtain  the  true  face,  or 
elevation  of  the  roof  as  it  would  appear  if  tipped  or 
swung  into  a  vertical  position,  for  the  purpose  of  get- 
ting the  correct  angle  at  A'  B1  F'. 

To  do  this  reproduce  the  section  of  mansard  and 
bed  molding  as  a  whole  at  a  convenient  point  below, 
but  so  turned  as  to  bring  the  faces  of  the  roof  into  a 
vertical  position,  maintaining  the  same  distance  be- 
tween the  points  A  and  B  as  shown  by  A"  and  B2. 
Project  lines  horizontally  to  the  left  from  this  section 
for  the  true  face,  marking  the  lines  from  the  points 
A2  and  B2.  From  A  of  the  original  section  carry  a  line 
across  intersecting  the  line  A1  B'  at  the  point  A1. 
Next  drop  line  from  A1  and  B'  vertically  intersecting 
iiiK-s  of  corresponding  letter,  as  shown  by  the  dotted 


lines.  Then  A3  B3  F'  will  be  the  correct  angle  upon 
which  to  construct  the  corner  piece  and  develop  the 
miter  line  between  the  hip  molding  and  the  bed  mold- 
ing of  the  deck  cornice. 

The  next  step  will  be  to  obtain  a  correct  section 
of  the  hip  molding  upon  a  line  at  right  angles  to  the 
line  of  the  hip.  To  do  this  it  is  necessary  to  first 
construct  a  diagonal  section  through  the  hip.  At  any 
convenient  place  lay  off  a  plan  of  the  angle  of  the  roof, 
as  shown  by  D1  F  D"  in  Fig.  407,  and  through  this  angle 
draw  a  plan  of  the  hip,  as  shown  by  F  K.  From  D1  erect 
a  line  perpendicular  to  F  D',  as  D1  C',  in  length  equal 
to  D  C  of  the  section.  Through  C",  parallel  to  D'  F, 
draw  C2  K,  producing  it  until  it  cuts  the  line  repre- 
senting the  plan  of  the  hip.  From  the  points  F  and  K 
in  the  lines  representing  the  plan  of  the  hip  erect  per- 
pendiculars, as  shown  by  F  L  and  K  C3.  Draw  L  C3 
parallel  to  F  K,  as  shown  at  the  base  line  of  a  diag- 
onal section.  From  C3  erect  a  perpendicular,  C3  E1,  in 
length  equal  to  C  E  of  the  original  section.  Connect 
E1  L.  Then  L  C3  E'  will  be  a  diagonal  section  of  a 
portion  of  the  roof,  and  L  E1  will  be  the  length  of  the 
hip  through  that  portion.  At  right  angles  to  L  E1 
draw  M  H1,  upon  which  to  construct  a  correct  section 
of  the  hi})  molding.  Take  any  point,  as  G  in  the  line 
F  D1,  at  convenience,  and  from  it  erect  a  perpendicular 


218 


The  New  Metal  Worker  Pattern  Book. 


to  P  K,  cutting  F  K  in  the  point  H,  and  produce  it  '  will  be  obtained  by  which  the  angle  contained  between 
also  iintil   it  cuts  the  base  line  of  the  diagonal  section      the   facias   of    the    hip   molding   mav   be  determined. 


L  C3,  as  shown,  and  from  this  point  carry  it  parallel  to 
the  line  L  E1,  representing  the  pitch  of  the  hip,  until  it 


Therefore  from  II'  on  either  side  set  off  the  distance 
H  G  of  the  plan,   as  shown  by  G1  and  G2.      Through 


Fiij.  jfOfi.—Thr  Pattern  of  a  Hip  Molding  Upon  a  Right  Anyle  in  a  Mansard  Rnof,  Miteriny  Atj<ni,nt  «  Bed 


crosses  the  line  M  H1,  cutting  it  in  the  point  II1. 
Since  D1  F  D*  represents  the  angle  in  plan  over  which 
the  hip  molding  is  to  lit,  and  since  H  G  is  the  meas- 
urement across  that  angle,  if  the  distance  II  G  be  set 
off  from  H1  either  way  in  the  diagonal  section,  points 


these  points  draw  lines  representing  the  fascias  of  the 
hip  molding,  as  shown  by  O  (I1  and  ()  (i2.  Add  the 
fillets  and  draw  the  roll  according  to  given  dimensions, 
all  as  shown. 

In  the  true  face,  Fig.  4nii.   draw  a  half  section  of 


Pattern  Problems. 


219 


the  hip  molding  as  derived  from  Pig.  407,  as  shown. 
-M'  IF. corresponds  to  M  H'  of   the  diagonal   section. 


i. /    ' —    ;c-< 

1          Diagonal  Section 

' 


Fig.  40?.— Method  »f  Obtainim./  t ',,,;•<•<*  Cross  Section  of  Hip  in 
Fig.  #»;. 


Space  this  profile  into  any  convenient  number  of  partis 
in  the  usual  manner,  and  through  the  points  draw  lines 
parallel'  to  tjie  lines  of  the  hip  molding  indefinitely. 


Place  a  corresponding  portion  of  the  profile  of  the  hip 
molding  in  the  vertical  section,  as  shown,  in  which 
M'  IF  also  corresponds  to  II1  M  in  the  diagonal  section. 
Divide  this  section  into  the  same  number  of  equal 
parts,  and  through  the  points  draw  lines  upward  until 
they  intersect  with  the  profile  of  the  bed  molding,,  as 
shown  between  P"  and  Ba.  From  the  points  i  n  P2  B"  carry 
lines  horizontally,  intersjecting  the  lines  drawn  from 
the  profile  in  the  .true  face.  Then  a  line  traced 
through  these,  points  of  intersection  will  be  the  miter 
line  between'  the, hip  molding  and  the  bed  molding,  as 
seen. near  B'.rh  elevation. 

Fpy  'the  pattern  proceed  as  follows :  At  right 
angles  to -the  line  of  the  hip  molding,  as  shown  in  the 
true  face,  lay  off  a- -stretchout  of  the  hip  molding,  as 
shown  by  S  R,  through  the  points  in  which  draw  the 
usual  measuring  lines.  Place  the  T-square  at  right 
angles  to  the  lines  of  the  hip  molding,  and,  bringing 
it  successively  against  the  several  points  in  the  miter 
line,  as  shown  in  elevation,  cut  corresponding  measur- 
ing lines,  which  will  give  the  pattern  for  the  roll  and 
fillets,  as  shown  from  U  to  V.  In  like  manner  place 
the  T-square  against  the  point  X  in  the  true  face, 
which  is  the  point  of  junction  between  the  flange  of 
the  hip  molding  and-,  the  apron  of  the  bed  molding 
corresponding  to  points  9  and  10  of  the  profile,  and 
cut  the  corresponding  measuring  lines.  The  pattern 
is  then  completed  by  drawing  u  line  from  W  to  V  and 
T  to  U. 


' 


PROBLEM  103. 

Patterns  for  the  Top  and  Bottom  of  the  Hip  Bar  in  a  Skylight. 


In  the  upper  part  of  Fig.  408  is  sJiown  the  trans- 
verse section  of  a  skylight  in  which  A  B  represents  a 
portion  of  the  ventilator  or  finish  at  the  top,  and  C  D 
the  curb  or  finish  at  the  bott  >m.  The  section  also 
shows  the  side  elevation  of  a  ''  common"  bar  whose 
profile  is  at  F.  The  plan  immediately  below  shows  a 
corner  of  the  skylight  with  one  of  the  hip  bars,  H  K, 
the  patterns  for  which  are  required.  It  will  be  neces- 
sary first  to  sec  that  the  plan  is  correctly  projected 
from  the  elevation,  and  afterward  that  a  diagonal  ele- 
vation of  the  hip  bar  bo  obtained  from  this  plan,  be- 
fore the  correct  or  raked  profile  of  the  hip  bar  can  be 
obtained. 

'  •  -Draw,  a  duplicate  of  -normal  profile  F  with  its  cen- 
ter line  on  the  center  line  of  the  hip,  as  shown  at  F', 


as  a  means  of  obtaining  thejmefal  projection  of  all  its 
points,  numbering  corresponding  points  in  both  profiles 
the  same.  Number  the  intersections  of  all  the  points 
in  the  normal  profile  F  with  the  top  and  .bottom  of  the 
skylight  finish,  as  shown  by  the  small  figures  in  A  B 
and  C  G.  From  each  of  the  points  in  the  profile  F 
carry  lines  parallel  to  the  center  line  of  the  hip  in  either 
direction,  intersecting  lines  of  corresponding  number 
dropped  vertically  from-bo-tk  4-ho  -miters  of  the  trans- 
verse section  to  the  plan.  Lines  traeed  through  these 
points  of  intersection  will  give  the  miter  lines  at  top 
and  bottom  as  they  appear  in  plan. 

At  right  angles  to  the  lines  of  the  hip  carry  lines, 
as  shown,  by  means. of  which  to  construct  the  diagonal 
elevation.  Assume  anv  line,  as  E'  G',  as  the  base  or 


22C 


Tlte  New  Metal    Worker  Pattern  Book. 


horizontal  line  of  the  diagonal  elevation  representing 
E  G  of  the  section.     At  E'  erect  a  perpendicular  upon 


the    horizontal    molding  at  the   top  whose  profile  is 
shown  at  A  B,    it  will  be  found  most  convenient  to 


Fig.  408.— Plan  and  Section  of  a  Skylight  and  Patterns  for  the  Hip  Bar. 


which  to  obtain  the  hights  of  the  various  points  in  the 
upper  miter.     As  the  hip  bar  is  required  to  miter  with 


carry  all  the  points  of  the  upper  profile  to  the  vertical 
line  A  B,  as  shown  by  1,  2,  3',  4,  5  and  (51,  and  after- 


Pattern  Problems. 


221 


ward  to  transfer  them,  as  shown,  to  the  line  A'  B', 
keeping  the  perpendicular  hight  from  E1  to  B1  equal 
to  E  B.  From  all  the  points  in  A'  B'  carry  lines  hori- 
zontally— that  is,  parallel  to  E'  G' — to  the  right  indefi- 
nitely, as  shown.  These  lines  will  then  represent  a 
partial  elevation  of  the  top  molding  A  B  in  the 
diagonal  elevation.  Lines  from  each  of  the  points  in 
the  plan  of  the  upper  miter  at  II  may  now  be  carried 
parallel  to  II  E'  until  they  intersect  with  lines  of  cor- 
responding number  drawn  from  A'  B1.  Lines  connect- 
ing the  points  of  intersection  will  give  the  required 
miter  line  at  the  top  of  the  hip  bar. 

From  each  of  the  points  obtained  in  this  miter 
line  carry  lines  parallel  to  B'  G',  or  the  rake  of  the  hip 
bar,  and  intersect  them  with  lines  projected  parallel  to 
II  E'  from  the  lower  miter  in  plan  at  K.  Lines  con- 
necting these  points  of  intersection  will  give  the  re- 
quired miter  line  at  the  bottom  of  the  hip  bar. 

It  now  remains  only  to  obtain  the  correct  profile 
of  the  hip  bar  before  a  stretchout  can  be  obtained. 
To  accomplish  this,  draw  any  line  cutting  the  lines  of 
the  hip  bar  in  the  diagonal  elevation  at  right  angles,'as 
shown  at  R.  Upon  this  line,  and  above  or  below  the 
hip  bar,  as  shown  at  F",  draw  a  duplicate  of  the  normal 
profile  F,  from  the  points  in  which  carry  lines  at  right 
angles  to  the  hip  bar,  cutting  lines  of  corresponding 
number  in  the  same.  Then  lines  connecting  the  points 
'  of  intersection  will  give  the  raked  profile,  as  shown  at  R. 

On  account  of  limited  space  the  important  details 


in  Fig.  108  are  necessarily  small,  but  great  care  has 
been  taken  in  the  preparation  of  the  drawing,  and  all 
the  points  in  the  several  views  of  both  miters  have 
been  carefully  numbered,  so  that  the  reader  will  have 
no  difficulty  in  following  out  the  various  intersections 
from  start  to  finish.  The  profile  and  the  two  miter 
lines  now  being  in  readiness,  the  pattern  may  be  de- 
veloped in  the  usual  manner,  as  follows  :  Upon  any 
line  drawn  at  right  angles  to  the  hip  bar,  as  L  M,  lay 
off  a  stretchout  of  the  profile  R,  as  shown  by  the  small 
figures,  through  which  draw  the  measuring  lines. 
Keeping  the  blade  of  the  I  -square  parallel  with  L  M, 
bring  it  successively  against  the  points  of  intersection 
previously  obtained  in  the  upper  and  lower  miters  and 
cut  corresponding  measuring  lines.  Then  lines  traced 
through  the  various  points  of  intersection,  as  shown  by 
N  0  and  P  Q,  will  constitute  the  required  patterns. 

It  may  be  noticed  that  while  most  of  the  points 
from  the  normal  profile  ¥  come  squarely  against  the 
inner  beveled  surface  of  the  curb  G,  the  points  1  and 
.  2,  representing  the  vertical  portion  of  the  bar,  pass 
over  the  curb  to  a  point  beyond.  The  line  from  point 
2,  therefore,  intersects  at  both  2  and  2',  which  points 
are  duly  carried  through  the  views  of  this  miter  at  K 
and  G1  and  finally  into  the  pattern,  as  shown ;  from 
which  it  may  be  seen  that  the  miter  pattern  may  be 
cut  as  shown  by  the  solid  line  from  P  to  Q,  or  that 
portion  from  point  2  to  3  may  be  cut  as  shown  by  the 
dotted  line. 


PROBLEM  104. 

Pattern  for  the  Top  of  a  Jack  Bar  in  a  Skylight. 


The  jack  bar  in  a  skylight  is  the  same  as  the 
"  common  "  bar  in  respect  to  its  profile,  and  the  miter 
at  its  lower  end  with  the  curb.  At  its  upper  end, 
however,  it  is  required  to  miter  against  the  side  of  the 
hip  bar.  instead  of  against  the  upper  finish  of  the  sky- 
light. As  the  hip  bar  occupies  an  oblique  position 
with  reference  to  the  jack  bar,  it  is  evident  that  a 
perfect  miter  between  the  two  could  not  be  effected 
without  a  modification  or  raking  of  the  profile  of  the 
hip  bar,  all  of  which  has  been  demonstrated,  in  the 
preceding  problem. 

It  may  be  here  remarked  that  the  raking  of  the 
profile  of  the  hip  bar  is  done  not  so  much  to  affect  a 
perfect  joint  with  the  top  finish  as  to  make  a  perfect  j 


miter  with  the  jack  bar,  or  what  is  the  same  thing, 
that  the  surfaces  indicated  by  2  3  of  the  profile  of  the 
hip  bar  in  Fig.  408  shall  lie  in  the  same  plane  with 
that  portion  of  the  profile  of  the  jack  bar.  However, 
as  the  raked  hip  bar  presents  exactly  the  same  appear- 
ance when  viewed  in  plan  as  a  bar  of  normal  profile, 
it  will  not  be  really  necessary,  so  far  as  the  miter  cut 
on  the  jack  bar  is  concerned,  to  perform  the  raking 
operation. 

In  Fig.  409  is  shown  a  sectional  and  a  plan  view 
of  a  portion  of  a  skylight  containing  the  miter  above 
referred  to.  The  normal  profile  of  the  jack  bar  shown 
at  F  and  F'  is  not  exactly  the  same  in  its  proportions 
as  that  of  the  preceding  problem,  but  possesses  the 


222 


77/e  Xvw  Mi'lul    Worker  Pattern  Book. 


same  general  features.  The  view  of  the  bar  given  in 
the  section  from  A  to  15  represents  an  oblique  eleva- 
tion of  that  side  of  ihe  hip  bar  which  is  toward  the 
jack  bar.  From  B  to  D  the  view  shows  the  side  of  the 
jack  bar,  while  beyond  D  is  shown  a  continuation  of 
the  full  hip  bar  with  its  profile  correctly  placed  in  po- 
sition at  F". 

The  first  step  before  the  pattern  can  be  laid  out 
is  to  obtain  a  correct  intersection  of  the  points  in  the 
plan,  as  at  B1,  and  afterward  an  elevation  of  the  same, 
as  shown  at  B.  Draw  u  normal  profile  of  the  jack  bar 
in  correct  position  in  the  plan,  as  shown  at  F'.  Al>n 
place  a  profile  of  the  hip  bar  in  the  plan  of  the  same, 
as  shown  at  F:.  As  only  the  lateral  projection  of  the 
[mints  are  here  made  use  of  a  normal  profile  will 
answer  as  well  as  the  raked  profile  shown,  as  above 
intimated.  Number  all  the  points  in  both  profiles 
Correspondingly,  and  from  the  points  in  each  carry 
lines  respective! v  parallel  to  their  plans,  intersecting 
as  shown  at  B'.  From  the  points  of  intersection  of 
like  numbers  erect  lines  vertically  into  the  sectional 
view,  cutting  lines  of  corresponding  number  drawn 
from  the  points  in  the  profile  F  parallel  to  the  lines  of 
Uhe  rake,  as  shown  near  B.  It  will'  be  seen  that  both 
sides  of  the  profile  F'  intersect  with  one  side  of  the 
profile  F".  both  sets  of  intersection  being  numbered 
alike,  as  I1,  ~2',  3',  etc.  This  gives  rise  to  two  miter 
lines  at  B  in  the  sectional  view.  The  line  correspond- 
ing to  the  intersections  on  the  upper  side  of  the  jack 
bar  are  here  numbered  1",  2",  '.'>',  etc.,  while  those 
points  belonging  exclusively  to  the  lower  intersection 
are  numbered  33,  ^  and  it;i. 

A  stretchout  of  the  normal  profile  F  mav  now  be 
laid  off  on  anv  line,  as  (i  II.  drawn  at  right  angles  to 
the  elevation  of  the  jack  bar,  through  which  the  usual 
measuring  lines  are  drawn.  Now  place  the  blade  of 
the  T-sqnare  parallel  to  G  H,  and,  bringing  it  against 
the  various  points  in  the  two  miter  lines  above  de- 
scribed, cut  corresponding  measuring  lines,  carrying  the 
points  from  the  upper  miter  line  into  one  side  of  the 
pattern  and  those  from  the  lower  one  into  the  other 
side;  then  lines  connecting  the  points  of  intersection, 
as  shown  from  K  to  L,  will  constitute  the  required 
miter  cut. 

As  it  is  desirable  to  cut  the  miter  on  the  jack  bar 
so  as  to  fit  o.ver  the  hip  bar  (that  is,  so  as  not  to  cut  the 
hip  bar  at  all)  and  in  order  to  prevent  the  surface  from 
4  to  5  of  the  jack  bar  from  hipping  on  to  a  like  por- 
tion of  the  hip  bar,  as  shown  between  the  points  4',  5" 
and  x  in  the  plan,  the  line  from  point  4  of  F1  is  al- 


lowed to  intersect  with  the  line  from  5  of  F3,  as  shown 
at  x,  which  point  is  carried  into  the  sectional  view  and 
thence  into  the  pattern,  where  it  intersects  with  lines  4, 
as  shown  by  x,  so  that  the  cut  in  the  pattern  is  from 
x  to  5  instead  of  from  4  to  5.  For  the  same  reason, 


Fig.  409. — Section  and  Plan  of  Miter  at  the  Top  of  the  Jack  Bar  in 
a  Skylight  and  Pattern  of  the  Same. 


if  it  is  desired  to  prevent  the  surfaces  2  3  from  over- 
lapping the  line  from  2  of  F1  may  be  intersected 
with  3  from  F2,  as  shown  at  ?/,  and  carried  into  the 
pattern,  as  shown,  producing  the  cuts  in  the  pattern 
shown  by  the  dotted  lines  3  y  in  the  place  of  those 
shown  from  3  to  2. 


I'lttti'.rn   Problems. 


223 


PROBLEM   105. 

The  Pattern  of  a  Hip  Mold  Upon  an  Octagon  Angle  in  a  Mansard  Roof,  Mitering  Against  a  Bed 

Molding  of    Corresponding   Profile. 


This  problem,  like  many  others  pertaining  to  man- 
sard roofs,  m;iv  reach  the  pattern  cutter  in  drawings 
either  more  or  less  accurate,  and  in  diil'erent  stages  of 
completion.  Certain  facts,  however — vi/.,  the  profiles 
of  the  moldings,  the  pitch  of  the  roof  and  the  angle  in 
plan — must  be  known  before  the  work  can  be  accom- 
plished ;  but  with  these  given  the  pattern  cutter  will 
have  no  difficulty  in  drawing  such  elevations  as  are 
necessary  to  produce  the  required  patterns. 

In  Fig.  410,  let  A  B  C  D  be  the  given  section  of 
the  mansard  trimming  shown,  A  C  the  profile  of 
the  bed  molding  and  apron,  and  B  D  E  the  pitch  of 
the  roof.  According  to  the  statement  of  the  problem 
above  the  angle  of  the  plan  is  octagonal ;  it  might  be 
a  special  angle  cither  greater  or  less  than  that  of  an 
octagon,  but  the  principle  involved  and  the  operation 
of  cutting  the  patterns  would  be  the  same.  As  in  all 
other  problems  connected  with  mansard  trimmings,  the 
first  requisite  is  an  elevation  of  the  "true  face,"  in 
order  to  obtain  the  correct  angle  between  the  bed  mold 
ing  and  the  hip  molding.  A  normal  elevation,  such 
as  is  likely  to  be  met  with  in  the  architect's  'drawings, 
is  shown  in  the  engraving  at  the  left  of  the  section, 
merely  for  purposes  of  design.  In  obtaining  the  true 
face,  shown  below,  it  is  best  to  use  the  section  and  plan 
onlv.  Therefore,  redraw  the  section  as  shown  im- 
mediately below  it,  placing  the  line  of  the  roof  in  a  ver- 
tical position,  all  as  shown.  From  all  the  points  of  this 
section  lines  may  now  be  projected  horizontally  to  the 
left,  as  the  first  step  in  developing  the  required  true 
face.  Immediately  above  the  space  allotted  to  the 
elevation  draw  a  plan  of  the  horizontal  angle,  as  shown 
by  I  E'  K.  As  it  will  be  impracticable  to  include  the 
entire  profile  of  the  roof  in  the  drawings,  some  point 
must  be  assumed  at  a  convenient  distance  below'  the 
bed  mold,  as  D,  from  which  to  measure  hight  and  pro- 
jection, which  locate  also  in  the  section  below,  as  shown 
at  D',  making  B1  D'  equal  to  B  D.  From  A  draw  a 
line  at  right  angles  to  the  line  of  the  roof,  meeting  it 
at  B,  which  point  may  be  assumed  for  convenience  as 
the  upper  limit  of  that  part  of  the  roof  under  consider- 
ation. Now,  from  the  point  B  drop  a  vertical  line, 
*which  intersect  with  one  drawn  horizontally  from  D,  as 
shown  at  E;  then  1)  E  will  represent  the  projection. 


From  the  lines  I  K'  aud  F'  K.  upon  lines  at  right 
angles  to  each,  set  oil'  the  projection  D  E,  as  shown  at 
I  F  and  K  II;  through  the  points  F  and  11  draw  linos 
parallel  to  the  first  lines,  meeting  in  G;  then  a  line  from 
<!  to  II  will  represent  the  plan  of  the  angle  or  hip  of 
that  portion  of  the  roof  of  which  B  D  is  the  profile. 
N"o\v,  to  complete  the  true  face  of  that  part  of  the 
roof  drop  a  line  from  the  point  E'  intersecting  the  line 
fro.n  B1  at  L,  and  one  from  G  intersecting  the  one  from 
D1  at  M ;  then  the  angle  B1  L  M  will  be  the  correct 
angle  of  the  miter  between  the  bed  mold  and  the  hip 
mold. 

As  in  Problems  lUl  and  102  preceding,  it  will  next 
be  necessary  to  obtain  a  correct  section  of  the  hip  mold 
on  a  line  at  right  angles  to  the  line  of  the  hip.  To  avoid 
confusion  of  lines,  this  operation  is  shown  in  Fig.  411, 
in  which  E"  G1,  the  base  line,  is  made  equal  to  E1  G  of 
the  plan  in  the  previous  figure.  At  the  point  Ea  erect 
a  perpendicular,  making  it  equal  in  hight  to  B  E  of  the 
sectional  view.  Connect  B"  with  G',  which  will  give 
the  correct  angle  of  the  hip  of  the  roof.  As  a  means 
of  constructing  a  correct  section  at  right  angles  to  this 
line,  assume  any  two  points  on  the  original  plan,  as  N 
and  0,  equidistant  from  G  and  connect  them  by  a 
straight  line,  cutting  the  angle  or  hip  line  in  P.  Set  off 
from  G1  on  the  line  G'  E"  of  Fig.  411  a  distance  equal 
to  G  P  of  the  plan,  as  shown  at  1",  from  which  draw  a 
line  parallel  to  the  hip  G'  W.  Next  intersect  these 
two  lines  by  another  at  right  angles  at  any  convenient 
point,  as  shown  by  P2  Q.  From  the  point  P"  set  off  the 
distances  P3  O1  and  P"  N',  making  them  equal  to  P  0 
and  PN.  Connect  the  points  O1  and  N1  with  R,  which 
is  the  intersection  of  P"  Q  with  the  hip  line ;  then  the 
angle  O'  R  N1  will  be  a  correct  section  of  thereof  upon 
the  line  P3  Q  or  upon  any  line  cutting  the  hip  at  right 
angles,  upon  which  the  finished  profile  of  the  hip  mold 
may  now  be  constructed ;  as  follows:  Set  off  the  pro- 
jections of  the  fascia  and  fillet  as  given  in  the  sectional 
view,  Fig.  410,  from  the  lines  R  O1  and  R  N',  contin- 
uing their  lines  to  the  center  line  P"  Q.  From  the  inter- 
section S  as  a  center,  with  a  radius  of  the  bed  mold, 
describe  the  roll. 

As  stipulated  in  the  statement  of  this  problem, 
the  profiles  of  the  bed  mold  aiul  hip  mold  are  to  cor- 


224 


Xew  Metal    Worker  Pattern    Book. 


Fig.  ill. -Diagonal 
Section  of  flip. 


Fig.  410.— The  Pattern  of  a  Hip  Molding  Upon  an  Octagon  Angle,  Miteriiig  Against  a  Heil  MMituj  of  Corresponding  Profile. 


Pattern  Problems. 


225 


respond.  By  this  it  is  understood  that  the  curves  of 
their  molded  surfaces  are  alike  and  struck  with  the  same 
radius,  and  so  placed  as  to  member  or  miter. 

As  the  curve  of  the  bed  mold  is  only  a  quarter 
circle,  while  that  of  the  hip  mold  is  nearly  three- 
quarters  of  a  circle,  it  will  be  seen  that  the  quarter 
circles  in  each  half  of  the  hip  mold  next  adjacent  to  the 
fascias  and  iillets  will  miter  with  the  arms  of  the  bed 
mold  on  either  side  of  the  miter,  and  that  a  small  space 
in  the  middle  of  the  roll  will  remain  between  them, 
which  must  be  mitered  against  the  planceer,  and  the 
object  of  the  operation  shown  in  Fig.  411  is  to  deter- 
mine exactly  what  this  space  is.  The  dotted  lines 
from  S  to  the  points  10  drawn  at  right  angles 
to  K  (.)'  and  R  N'  show  the  limit  of  the  quarter 
circles  or  the  parts  that  must  miter  with  the  bed  mold, 
while  the  space  between  them  (10  to  10)  shows  the 
part  that  must  miter  against  the  planceer.  It  might  be 
supposed  that  the  angle  between  the  fascias  of  the  hip 
mold,  to  fit  over  the  angle  of  a  mansard  which  is  octag.- 
onal  in  plan,  would  be  octagonal,  but  the  demonstra- 
tion shows  that  while  the  angle  N  GO  of  tjie  plan, 
Fig.  410,  is  that  of  an  octagon,  the  angle  N1  R  O', 
Fig.  411,  is  greater,  because  the  distance  N'  0'  is  equal 
to  X  O,  while  the  distance  R  PJ  is  less  than  G  P,  R  P3 
being  at  right  angles  to  the  line  of  the  hip  and  G1  P' 
being  oblique  to  it. 

The  true  face,  Fig.  410,  may  now  be  completed, 
as  follows :  Upon  any  line,  as  S1  T,  drawn  at 
right  angles  to  L  M,  representing  the  face  of  the  roof, 
draw  a  duplicate  of  one-half  the  profile  of  the  hip  mold 
obtained  in  Fig.  411,  placing  the  point  S  upon  the  line 
L  M,  as  shown.  Lines  drawn  through  the  angles  of 
this  profile  parallel  to  L  M  will  intersect  with  lines 
from  corresponding  points  from  the  profile  A1  C',  pre- 
viously drawn,  giving  the  miter  line  .1  X  and  com- 
pleting the  elevation  of  the  true  face. 


I 'pon  any  line  at  right  angles  to  the  line  L  M,  as 
U  V,  lay  off  a  stretchout  of  the  complete  hip  mold  as 
obtained  from  the  half  profile  S'  T,  through  which 
draw  measuring  lines  as  usual.  Drop  lines  from  the 
points  from  1  to  10  of  the  profile,  parallel  to  L  M,  cut- 
ting the  miter  line;  then,  with  the  T-square  placed  at 
right  angles  to  L  M  and  brought  successively  against  the 
points  in  J  X,  intersect  them  with  lines  of  corresponding 
number  in  the  stretchout;  then  lines  traced  through  the 
points  of  intersection,  shown  by  d  I  and  a  c,  will  give 
the  pattern  for  that  nart  of  the  profile  from  1  up  to  the 
point  10.  The  pattern  ci  that  portion  of  the  roll  which 
miters  against  the  planceer  muso  be  obtained  from  the 
diagonal  section  of  the  hip.  From  points  10,  11  and 
12  in  Fig.  411  carry  lines  parallel  to  G'  B*  intersecting 
the  line  of  the  planceer,  as  shown  at  W.  It  is  only 
necessary  to  ascertain  how  much  shorter  the  lines  11 
and  12  are  than  the  line  10,  and  then  to  transfer  these 
distances  to  the  pattern.  This  can  be  done  by  drop- 
ping lines  from  the  intersection  of  points  11  and  12 
with  the  planceer,  in  Fig.  411,  at  right  angles  to  G'  B3, 
cutting  line  10.  These  distances  can  then  be  trans- 
ferred to  line  10  of  the  pattern,  Fig.  410,  measuring 
down  from  the  point  10  of  pattern  already  obtained, 
after  which  they  may  be  carried  parallel  to  U  V  into 
the  measuring  lines  11  and  12,  thus  completing  the 
pattern. 

This  portion  of  the  work  is  necessarily  very  minute 
in  the  drawing,  but  it  will  be  easily  seen,  in  applying 
the  principle  to  other  similar  cases,  that  if  the  angle  of 
the  plan  I  E1  K  were  less  than  that  shown,  for 
instance,  if  it  were  a  right  or  an  acute  angle,  a 
greater  distance  or  more  points  would  occur  between 
the  points. 10  and  10,  and  further,  that  if  the  angle  of 
the  roof  were  less  steep  a  greater  curve  or  dip 
would  occur  between  those  points  (a  to  i)  of  the 
pattern. 


PROBLEM  106. 

The  Pattern  of  a  Hip  Molding  Upon  an  Octagon  Angle  of  a  Mansard  Roof,  Mitsring  Upon  an  Inclined 

Wash  at  the  Bottom. 


In  Fig.  412,  let  D  B  of  the  section  represent  the 
wash  surmounting  the  base  molding  at  the  foot  of  a 
mansard  roof,  the  inclination  of  the  roof  being  shown 
by  B  A.  The  plan  of  the  angle  of  the  roof  B2  K  B2, 
as  specified,  is  that  of  an  octagon,  but  so  far  as  prin- 
ciple and  method  are  concerned,  it  may  be  any  angle 


whatever.  The  profile  of  the  hip  mold  as  given  in  the 
original  drawings  will  most  likely  be  drawn  as  fitting 
over  an  octagonal  angle — that  is,  over  the  angle  as  given 
in  the  plan  of  the  building.  As  explained  in  the 
problem  preceding  this,  a  section  through  the  angle  of 
the  roof  at  right  angles  to  the  line  of  the  hip  must  be 


226 


Tlie  New  Metal    Worker  Pattern   Book. 


Ks     |?  P3 

Fig.  U1S  —  Diagonal 
Section  of  Hip. 


,  Ri 

f  D 


G"  H* 

2— The  Pattern  of  a  Hip  Molding  Upon  an  Octagon  Angle,  Mitering  Upon  an  Inclined  Wash  at  the  Bottom. 


Pattern  Problems. 


227 


obtained,  to  which  the  profile  of  the  hip  mold  must  be 
adjusted  before  going  ahead.  The  difference  between 
such  a  section  and  the  angle  in  plan  may  seem  trilling, 
but  will  be  found  to  increase  as  the  pitch  of  the  roof 
decreases,  and  in  a  low  hip  roof  will  be  found  to  be 
considerable.  Hence  the  original  detail  of  the  hip 
mold  must  be  accepted  only  so  far  as  it  gives  width 
and  depth  of  fascias  and  fillets,  and  diameter  or  radius 
of  the  roll,  while  the  angle  between  the  faseias  must 
be  adjusted  to  the  true  section  across  the  hip  as  above 
stated.  The  method  of  doing  this  is  shown  in  Fig. 
!13,  and  the  principles  involved  therein  arc  explained 
in  the  previous  problem  m  connection  with  Fig.  411, 
and  need  not,  therefore,  be  repeated  here. 

The  first  operation  will  consist  in  obtaining  the 
"true  face"  of  the  roof  in  the  usual  manner,  viz.: 
Assume  any  point  upon  the  section  of  the  roof,  as  A, 
at  a  convenient  distance  above  the  base,  as  a  point 
from  which  to  measure  hight  and  projection.  Redraw 
the  section  of  the  roof  immediately  below  the  first  one, 
placing  it  in  a  vertical  position  and  locating  thereon 
t'.ie  point  A,  as  shown  by  A1.  From  the  points  A1,  B1 
and  D1  project  lines  horizontally  to  the  left,  thus  ob- 
taining all  the  hights  in  the  true  face.  It  will  be 
necessary  next  to  complete  the  plan,  to  do  which  first 
i  litain  the  projection  of  the  points  in  the  section  upon 
any  horizontal  line,  as  the  one  drawn  through  B,  which 
can  be  done  by  dropping  vertical  lines  from  the  points 
A  and  D,  cutting  it  as  shown  at  I  and  C.  Assuming 
the  line  B2  K  B2  of  the  plan  to  represent  the  point  B 
of  the  section,  set  off  upon  any  lines  at  right  angles  to 
the  lines  B3  K  these  projections — that  is,  make  B"  I' 
equal  to  B  I,  and  B2  C1  equal  to  B  C.  Through 
these  points  draw  lines  parallel  to  B"  K,  intersecting 
and  forming  the  line  P  G,  whicn  is  the  plan  of  the 
angle  over  which  the  hip  mold  is  required  to  fit.  From 
the  points  P,  K  and  G,  which  represent  upon  the  angle 
of  the  roof  the  points  A,  B  and  D  of  the  section,  drop 
lines  vertically  into  the  true  face  intersecting  the 


horizontal  lines  previously  drawn  from  A',  B'  and  D1, 
as  shown ;  then  P'  K1  I2  will  be  the  correct  angle  at 
which  to  construct  the  miter  of  the  half  of  the  hip 
mold  belonging  to  this  face  of  the  roof,  and  K'  G' 
IP  P  will  represent  a  corresponding  elevation  of  the 
wash. 

The  elevation  01  the  true  face  may  now  be  com- 
pleted by  placing  one-half  the  profile  of  the  hip  in 
correct  position — that  is,  with  its  base  line  or  fascia  at 
right  angles  to  the  hip  line  P'  K',  the  point  R  coming 
on  the  line.  Through  the  points  Y,  S  and  T  project 
lines  parallel  to  the  hip  line.  To  show  the  intersection 
of  the  hip  mold  with  the  wash,  first  place  a  duplicate 
of  the  half  profile  of  hip  mold  in  the  sectional  view,  as 
shown  by  Y1  R1  T' ;  then  divide  the  curved  portion  of 
both  profiles  into  the  same  number  of  equal  spaces  and 
number  all  the  points  correspondingly,  as  shown. 
From  these  points  drop  lines  downward  parallel  with 
the  lines  of  the  respective  views,  those  in  the  sectional 
view  cutting  the  line  of  the  wash  B1  D1.  From  these 
points  of  intersection  carry  lines  horizontally,  intersect- 
ing the  lines  dropped  from  the  profile  Y  S  T.  Then  a 
line  traced  through  these  points  of  intersection,  as 
shown  by  Y2  S3  T2,  will  be  the  miter  line  formed  by 
the  junction  of  the  hip  molding  with  the  wash.  At 
right  angles  to  the  line  of  the  hip  molding  in  the  true 
face  lay  off  a  complete  stretchout  of  the  hip  molding, 
;  as  shown  by  U  V.  Through  the  points  in  it  draw 
measuring  lines  in  the  usual  manner.  Place  the  T- 
square  parallel  to  this  stretchout,  or,  what  is  the  same, 
at  right  angles  to  the  line  of  the  hip  molding,  as  shown 
in  true  face,  and,  bringing  it  successively  against  the 
points  in  the  miter  line  Y2  S2  T2,  cut  the  corresponding 
measuring  lines.  Then  a  line  traced  through  these 
points  of  intersection,  as  shown  from  W  to  Z,  will  be 
the  cut  to  fit  the  bottom  of  the  hip  molding. 

The  normal  elevation  may  be  completed,  if  desired, 
by  means  of  projections  from  the  plan  and  the  section, 
as  shown. 


Pattern  for  a   Hip 


PROBLEM  107. 

Molding:  Mltering    Against  the  Planceer  of  a  Deck  Cornice  on  a  Mansard  Roof 
Which  is  Square  at  the   Eaves  and  Octagon  at  the  Top. 


In  Fig.  414  is  shown  the  method  of  obtaining  the 
in  it  or  against  the  planceer  of  a  deck  cornice  formed  by 
the  molding  covering  a  hip,  which  occurs  between  the 


main  roof  and  that  part  wnich  forms  the  transition 
from  a  square  at  the  base  to  an  octagon  shape  at  the 
top.  The  roof  is  of  the  character  sometimes  employed 


228 


Tliv  Xcw  Mdal    \Ywlxr  Pattern  Bouk, 


upon  towers  which  are  square  in  a  portion  of  their  bight 
and  octagon  in  another  portion,  the  transition  from 
square  to  octagon  occurring  in  the  roof.  The  hip 
molding  in  question  covers  what  may  be  called  a  transi- 
tion hip,  being  a  diagonal  line  starting  from  one  of  the 
corners  of  the  square  part  and  ending  at  one  of  the 
corners  of  the  octagon  above.  A  carefully  drawn  plan, 
together  with  a  section  through  one  of  the  sides  of  the 
roof,  giving  the  pitch,  will  be  the  first  requisites  to 
solving  the  problem,  both  of  which  are  shown  in  the 
engraving.  The  first  operation  will  be  the  construc- 
tion of  a  section  upon  the  line  of  the  hip,  which  may 
be  done  as  follows :  Assume  any  point,  as  A,  in 
the  section  of  the  roof  from  which  to  measure 
bight  and  projection.  If  a  horizontal  line  from  A 
and  a  vertical  line  from  the  top  of  the  roof  surface 
B  be  intersected  in  C,  then  B  C  will  represent 
the  bight  and  A  C  the  projection  of  the  part  of 
the  roof  assumed.  Set  off  the  projection  C  A  at 
right  angles  to  the  top  line  of  the  plan  C'  E,  as  shown 
by  C1  A1,  and  carry  a  line  through  A'  parallel  to  C'  E 
till  it  cuts  the  plan  of  hip  E  Q  at  D ;  then  D  E  will 
form  the  base  and  B  C  the  bight  of  the  required  sec- 
tion, which  may  be  obtained  for  convenience  by  lines 
projected  from  D  E  at  right  angles,  all  as  shown.  The 
line  B'  D"  then  represents  the  real  angle  at  which 
the  hip  mold  meets  the  planceer  or  level  line  at 
the  top. 

The  next  operation  will  consist  in  obtaining  a  cor- 
rect section  of  the  hip  mold  from  the  data  given  and  in 
placing  it  in  correct  position  in  the  diagonal  section. 
Take  any  point,  G,  in  the  plan  at  a  convenient  distance 
from  the  angle  W  D  A1.  Set  off  G1  at  the  same  dis- 
tance from  the  angle  on  the  opposite  side.  From  the 
points  G  and  G'  carry  lines  at  right  angles  to  and  cut- 
ting D3  C3  in  the  points  H"  and  O3,  and  from  these 
points  carry  them  parallel  with  the  line  D2  B'  indefi- 
nitely. At  right  angles  to  D'  B'  draw  a  line,  as  shown 
by  Z  II1,  intersecting  with  the  lines  last  drawn  in  the 
points  H1  and  0.  From  H',  along  the  line  H'  PF,  set 
off  a  distance  equal  to  II  G  of  the  plan,  and  from  0, 
in  the  line  Z  H1,  set  off  a  distance  equal  toO'  G1  of  the 
plan,  as  shown  by  0  G".  Connect  the  intersection  of 
Z  H1  and  D'  B1  with  the  points  G3  and  G3,  which  will 
give  the  correct  section  through  the  angle  of  the  roof. 
Having  thus  determined  the  angle  of  the  hip  molding 
finish,  a  representation  of  it  is  indicated  in  the  drawing 


by  adding  the  flanges  and  the  roll.  Since  the  mittr 
required  is  the  junction  between  the  hip  molding,  the 
profile  of  which  has  just  been  drawn,  and  a  horizontal 
planceer,  the  remaining  step  in  the  development  of  the 
pattern  consists  simply  in  dividing  the  profile  into  any 
convenient  number  of  parts,  and  carrying  points  against 
the  line  of  the  planceer,  as  shown  near  B1,  and  thence 
carrying  them  across  to  the  stretchout,  as  indicated. 
It  is  evident,  however,  upon  inspection  of  the  eleva- 
tion, that  the  apron  or  fascia  strips  in  connection  with 
the  planceer  which  miter  with  the  flanges  of  the  hip 
molding  will  form  a  different  joint  upon  the  side  cur- 
responding  to  the  transition  piece  of  the  roof  than 
upon  the  side  corresponding  to  the  normal  pitch  of  the 
roof,  owing,  to  the  difference  in  pitch  of  these  two 
sides.  To  obtain  the  lines  for  this  rniter  an  additional 
section  must  be  constructed,  corresponding  to  a  center 
line  through  the  transition  piece,  as  shown  by  W  L  in 
plan.  Prolong  C3  D",  as  indicated,  in  the  direction  of 
W1,  and  lay  off  W  L',  equal  to  W  L  of  the  plan.  From 
L1  erect  a  perpendicular,  as  shown  by  L'  B%  equal  to 
C  B  of  the  original  section.  Connect  W  and  B3, 
against  the  face  of  which  draw  a  section  of  the  apron  or 
fascia  strip  belonging  to  the  planceer,  as  shown,  and  from 
the  points  in  it  carry  lines  parallel  to  B"  B'  until  they 
intersect  lines  drawn  from  the  flange  of  the  hip  molding 
lying  against  that  side  of  the  roof,  all  as  indicated  by 
U  X.  From  these  points  carry  lines,  cutting  corre- 
sponding lines  in  the  stretchout.  The  lines  of  the  fascia 
belonging  to  the  other  side  are  the  same  as  if  projected 
from  the  normal  section  at  B,  or  as  they  appear  in  the 
elevation.  Having  obtained  these  points  proceed  as 
follows :  At  right  angles  to  the  lines  of  the  molding  in 
the  diagonal  section  lay  off  the  stretchout  of  the  hip 
molding  S  T,  and  through  the  points  draw  the  usual 
measuring  lines,  as  shown.  Place  the  T-square  at  right 
angles  to  the  lines  of  the  molding,  or,  what  is  the  same, 
parallel  to  the  stretchout  line,  and,  bringing  it  success- 
ively against  the  points  formed  by  the  intersection  of 
the  lines  drawn  from  the  hip  molding  and  the  planceer 
line  B1,  cut  the  corresponding  measuring  lines,  as  shown. 
In  like  manner  bring  the  f -square  against  the  points  U 
and  X,  above  described,  and  Y  and  V,  points  corre- 
sponding with  the  opposite  side  of  the  hip  molding, 
and  cut  corresponding  lines.  Then  a  line  traced 
through  these  several  points  of  intersection,  as  shown 
by  U'  X'  Y1  V,  will  be  the  pattern  sought. 


Pattern   Problems. 


22!) 


SECTION 


fig.  414.— The  Pattern  for  a  Hip  Molding  Mitering  Against  the  Planceer  of  a  Deck  Cornice  on  a  Mansard  Roof  Which  is  Square  at  tht 

Eaves  and  Octagon  at  the  Top. 


230 


Tlie  New  Metal    Worker  Pattern  Book. 


PROBLEM  108. 

Patterns  for  a  Hip  Molding  Mitering  Against  the  Bed  Molding  of  a  Deck  Cornice   on  a  Mansard  Roof 

which  is  Square  at  the  Base  and  Octagonal  at  the  Top. 


The  problem  presented  in  Fig.  415  is  similar  to 
that  described  in  the  previous  problem,  with  the  dif- 
ference that  a  bed  molding  is  introduced  in  connection 
with  the  planceer  against  which  the  hip  molding  is  to 
be  mitered.  MEM1  represents  a  plan  of  the  roof  at 
the  top,  while  L  D  M»  represents  a  horizontal  line  at 
the  point  A  of  the  section,  assumed  at  convenience 
somewhere  between  the  top  and  the  bottom  for  the 
purpose  of  measurement.  The  intersection  of  the 
lines  M  L  and  E  D  prolonged  would  indicate  the  cor- 
ner of  the  building  at  the  bottom  of  the  roof,  the 
structure  being  square  at  the  base  and  octagonal  at 
the  top. 

The  first  step  in  the  development  of  the  pattern 
is  to  obtain  a  correct  section  of  the  roof  on  the  line  of 
one  of  the  hips.  Therefore,  at  any  convenient  point 
lay  off  E3  D3  of  Fig.  416  equal  to  D  E  of  the  plan. 
From  the  point  E3  erect  a  perpendicular,  E3  B",  in 
length  equal  to  C  B  of  the  section  of  the  roof.  Con 
nect  B"  and  D3,  which  will  be  the  pitch  of  the  hip 
corresponding  to  the  line  D  E  of  the  plan.  Since  the 
section  D3  E3  B"  has  been  constructed  away  from  and 
out  of  line  with  the  plan,  it  will  be  necessary  to  re- 
produce a  portion  of  the  plan  in  immediate  connection 
with  the  section,  as  shown  by  I'  II  A3  C".  This  can 
be  done  by  tracing,  or  any  means  most  convenient. 
From  the  point  II  in  this  plan  lay  off  on  either  arm 
the  points  I  and  I1,  equally  distant  from  it  and  con- 
veniently located  for  use  in  constructing  the  profile  of 
the  hip  molding.  From  the  points  I  and  I'  erect  per- 
pendiculars to  II  Cs,  cutting  it  in  the  points  K  and  0, 
which  prolong  until  they  meet  the  base  D3  E3  of  the 
diagonal  section,  from  which  points  carry  them  paral- 
lel to  the  inclined  line  D3  B'J  indefinitely.  At  right 
angles  to  the  inclined  line  D3  B2  draw  a  straight  line, 
O1  K1,  cutting  the  lines  last  described  in  the  points 
K'  and  0'.  From  K1,  measuring  back  on  the  line 
P  K1,  set  off  the  point  F,  making  the  distance  from 
K1  to  F  the  same  as  from  K  to  I  of  the  plan.  From 
O1  in  the  line  I1  I3  set  off  the  distance  O1  Is,  equal  to 
O  I'  of  the  plan.  From  these  points  I3  and  F  draw 
lines  meeting  the  line  O'  K'  at  the  point  of  its  inter- 


section with  the  line  D3  B".  Complete  the  profile  of 
the  hip  molding,  as  indicated,  laying  off  the  width  of 
the  fascias  on  these  lines,  adding  the  roll  and^dges. 

The  next  step  in  the  development  of  the  pattern 
is  to  draw  a  "  true  face  "  of  the  roof.  In  performing 
this  operation  it  matters  not  whether  the  actual  sur- 
face of  the  roof  be  used  or  the  surface  of  the  fascias. 
In  this  case  the  points  A  and  B  of  Fig.  415,  by  which 
the  depth  and  projections  of  the  pitch  are  measured, 
are  taken  on  the  surface  of  the  fascia.  For  the  true 
face  transfer  the  section  A  B  to  a  vertical  position, 
as  indicated  by  A'J  B',  Fig.  415,  in  connection  with 
which  the  bed  molding  against  which  the  hip  mold- 
ing is  to  miter  is  also  drawn,  as  shown.  From  the 
several  points  in  this  vertical  section  draw  horizontal 
lines,  which  intersect  by  vertical  lines  dropped  from  cor- 
responding points  in  plan.  Then  D5  E2  X  is  the  true 
face  of  that  part  of  the  roof  corresponding  to  D  E  M'  M* 
of  the  plan.  In  connection  with  the  vertical  section  just 
described,'  place  a  half  profile  of  the  hip  molding,  a 
true  section  of  which  has  been  obtained  by  the  process 
already  explained  in  Fig.  41G,  and  also  place  a  dupli- 
cate of  this  portion  of  the  profile  in  connection  with 
the  true  face.  Space  both  of  these  profiles  into  the 
same  number  of  parts,  and  from  the  several  points 
in  each  carry  lines  upward  parallel  respectively  to  the 
lines  of  the  views  in  which  they  appear;  the  lines 
from  the  profile  in  the  vertical  section  cutting  the  bed 
molding,  and  the  lines  from  the  profile  in  the  true  face 
being  continued  indefinitely.  From  the  points  of  in- 
tersection in  the  bed  molding  carry  lines  horizontally, 
intersecting  those  drawn  from  the  profile  in  connection 
with  the  true  face,  producing  the  miter  line,  as  shown 
by  E'. 

By  inspection  of  the  plan  where  a  portion  of  the 
bed  mold  is  shown  it  will  be  seen  that  the  miter  of  the 
bed  molding  around  the  octagon  at  E  is  regular — that 
is,  its  miter  line  does  not  coincide  with  the  line  of  the 
hip  D  E.  If  the  profile  of  the  bed  molding  in  the 
vertical  section,  and  also  the  profile  of  the  bed  mold- 
ing as  shown  in  the  plan,  be  divided  into  any  equal 
umber  of  parts,  points  may  be  dropped  from  the 
n 


Pattern  Problems. 


231 


Fig.  415.— Plan,  Elevation,  True  Face  and  Part  of  Pattern. 


Fig.  417.— True  Face  of  Octagonal  Side  and  Part  of  Pattern. 


Patterns  for  a  Hip  Molding  Miteriny  Against  the  Bed  Molding  of  a  Deck  Cornice  on  a   Mansard  Roof  which  is  Square  at  the  Base  and 

Octagonal  at  the  Top. 


232 


Tltc  New  Mdal    Worker  Pattern   iiook. 


profile  of  the  plan  on  to  the  miter  line  E,  and  thence 
carried  downward  and  intersected  with  horizontal  lines 
from  the  corresponding  points  of  the  bed  molding  in 
section,  also  shown  at  E2,  thus  giving  the  appearance  of 
the  miter  between  the  two  arms  of  the  bed  mold  behind 
their  intersection  with  the  hip  roll.  The  vertical 
lines  from  the  miter  E  of  the  plan  have  not  been  car- 
ried, in  the  engraving,  further  than  E1,  where  they  are 
intersected  with  lines  from  corresponding  points  from 
the  profile  at  B  of  the  elevation,  thus  showing  how 
the  operation  is  performed.  This  has  been  done  to 
avoid  a  confusion  of  lines  at  E2.  Having  obtained 
this  line  in  the  true  face,  the  point  where  it  crosses 
the  miter  line  between  the  hip  mold  and  bed  mold 
previously  obtained  at  E"  must  be  noted.  A  line  from 
this  point  of  intersection  must  then  be  carried  parallel 
to  the  line  of  the  molding  in  the  true  face,  back  to  the 
profile  of  the  hip,  and  there  marked,  as  shown  by  the 
figure  1\.  The  position  of  the  point  7£  should  now 
be  marked  upon  the  section  of  the  hip  molding  previ- 
ously obtained  at  O1  in  Fig.  416.  So  much  of  the 
profile  as  exists  between  1  and  7£  in  the  true  face  is 
used  in  obtaining  the  stretchout  of  this  part  of  the 
pattern.  The  remaining  portion  of  the  stay — namely, 
from  7£  to  14 — is  afterward  used  for  the  true  face  of 
the  octagonal  side  for  the  remainder  of  the  pattern. 

At  right  angles  to  the  line  of  the  molding  in  the 
true  face  lay  off  a  stretchout  equal  to  that  portion  of 
the  profile  thus  used,  as  shown  by  P  N,  through  the 
points  in  which  draw  measuring  lines  in  the  usual 
manner.  Place  the  T-square  at  right  angles  to  the 
lines  of  the  molding  in  the  true  face,  and,  bringing  it 
against  the  several  points  in  the  miter  line  between 
the  hip  and  bed  molding  at  E2, .  cut  corresponding 
measuring  lines  drawn  through  the  stretchout.  Then 
a  line  traced  through  these  points,  as  shown  by  S  T, 
will  be  the  miter  line  for  that  portion  of  the  pattern 
corresponding  to  the  part  of  the  profile  thus  used. 

For  the  other  half  of  the  hip  molding,  being  that 
portion  which  lies  on  the  face  of  the  transition  piece, 
another  operation  must  be  gone  through.  Construct  a 
section  of  the  roof  corresponding  to  the  line  F  G  in 
the  plan.  At  any  convenient  point  lay  off  F1  C1  in 
Fig.  415,  equal  in  length  to  F  G.  From  the  point  C1 
erect  a  perpendicular,  C1  B3,*in  length  equal  to  C  B  of 
the  section.  Connect  F1  and  B3.  Then  F1  B3  is  the 
length  of  the  transition  side  of  the  roof  through  that 


portion  corresponding  to  F  G  of  the  plan.  It  will  be 
well  to  add  to  this  at  B3  a  section  of  the  bed  mold  as 
it  appears  in  the  section  below,  thus  establishing  the 
true  relation  between  it  and  the  transition  side  of  the 
roof.  By  means  of  this  section  and  the  plan,  construct 
a  true  face  of  one-half  the  transition  side  of  the 
roof,  by  means  of  which  to  obtain  the  miter  of  the  re- 
maining portion  of  the  roll.  To  do  this  first  redraw 
the  section  B3  F',  placing  the  line  of  the  roof  in  a  ver- 
tical position,  as  shown  by  B4  F2,  Fig.  417,  from  the 
points  in  which  project  horizontal  lines,  as  shown  to 
the  right,  upon  each  of  which  set  off  from  an  assumed 
vertical  line  the  width  of  the  roof  as  given  in  the  plan. 
Thus  make  G'  E'  equal  to  G  E  of  the  plan,  and  V 
D4  equal  to  F  D.  Connect  D4  and  E'.  Then  G1  E4  D4 
F3  is  the  true  face  of  that  portion  of  the  roof  repre- 
sented by  G  E  D  F  in  the  plan. 

In  connection  with  the  vertical  section  just  de- 
scribed place  so  much  of  the  stay  as  was  not  used  for 
the  pattern  already  delineated,  and  in  the  elevation  of 
the  transitional  face  of  the  roof  place  a  corresponding 
portion  of  the  profile,  as  shown,  each  of  which  divide 
into  the  same  number  of  spaces.  From  the  points 
thus  obtained  carry  lines  parallel  to  the  lines  of  the 
respective  views  of  the  part,  those  in  the  vertical  sec- 
tion cutting  the  bed  molding,  and  those  in  the  eleva- 
tion being  produced  indefinitely.  From  the  points  in 
the  bed  molding  of  the  vertical  section  carry  lines 
horizontally,  intersecting  those  drawn  from  the  profile 
in  the  elevation,  thus  establishing  the  miter  line,  as 
indicated  at  E4.  At  right  angles  to  the  line  D4  E4  set 
off  a  stretchout  of  the  profile,  as  shown  by  R  P*, 
through  the  points  in  which  draw  the 'usual  measuring 
lines.  With  the  T-square  placed  parallel  to  this 
stretchout  line,  or,  what  is  the  same,  at  right  angles  to 
the  line  D4  E4,  and  being  brought  successively  against 
the  points  in  the  miter  line  at  E4,  cut  corresponding 
measuring  lines,  as  shown.  Points  also  are  to  be  car- 
ried across,  in  the  same  manner  as  described,  corre- 
sponding to  the  bottom  of  the  apron  or  fascia  strip  in 
connection  with  the  bed  molding.  Then  a  line  traml 
through  these  points,  as  indicated  by  the  line  drawn 
from  U  to  T',  will  be  the  pattern  of  the  other  half  of 
the  hip  molding.  By  joining  the  two  patterns  thus 
obtained  upon  the  dividing  line  of  the  stay,  correspond- 
ing to  P  T  of  the  first  piece  or  P2  T'  of  the  second 
piece,  the  pattern  will  be  contained  in  one  piece. 


(DUPLICATE   OF  PAGE  231.) 


B3 


Fig.  415.— Plan,  Elevation,  True  F*ce  and  Part  of  Pattern. 


Fig.  417.— True  Face  of  Octagonal  Site  anil  Part  of  Pattern. 


Patterns  for  a  Hip  Molding  M'dering  A&3.inst  the  Bed  Molding  of  a  Deck  Cornice  on  a   Mansard  Roof  which  is  Square  at  the  Base  and 

Octagonal  at  the  Top, 


234 


The  New  Metal   Worker  Pattern  Hook. 


-4«V A A' 


Fig.  418.~-The  Patterns  for  the.   Miter  at  the  Bottom  of  a  Hip  Molding  on  a  Mansard  Roof  Which  is  Octagon  at  the  Top  and  Square 

at  the  Bottom. 


Pattern  Problems. 


235 


correspond  to  the  point  8.  Locate  the  point  8  on  the 
first  section  of  the  hip  obtained  near  O,  as  shown,  and 
use  the  remainder  of  profile  8  to  14  for  another  opera- 
tion. Lay  off  a  stretchout  of  the  entire  profile  of  the 
hip  molding)  as  shown  by  W  V,  through  the  points  in 
which  draw  the  usual  measuring  lines.  With  the 


Fig.  419.— Section  and  Elevation  of  the   Miter  at  the  Bottom  of  a  Hip  Moldiwj 
on  a  Mansard  Roof  Which  is  Octagon  at  the  Top  and  Square  at  the  Bottom. 


Fig.  420.— Miter  between  the  Inner  Edges  of  the  Hip  Moldings  at  the  Bottom. 


T-square  placed  at  right  angles  to  the  lines  of  the  hip, 
as  shown  in  the  true  face,  and  brought  against  the 
points1  in  the  miter  line  S  T  TJ,  cut  so  many  of  the 
measuring  lines  drawn  through  the  stretchout  W  V  as 
correspond  to  those  points.  By  this  means  that  por- 
tion of  the  pattern  shown  by  S1  T'  IT  will  be  obtained. 


hip  mold  it  will  be  necessary  first  to  construct  a  true 
face  of  the  octagon  side  of  the  roof.  To  do  this, 
obtain  a  diagonal  section  of  the  roof  corresponding  to 
the  line  D  E  in  the  pjftn,  viz.  :  Lay  off  D3  E1  equal  to 
D  E  of  the  plan,  and  from  E'  erect  a  perpendicular, 
E1  A',  equal  to  C  A  of  the  section  in  Fig.  419.  Con- 
nect A'  and  D'.  Then  A"  D'  is  the 
length  of  the  diagonal  face  of  the  roof 
measured  on  the  line  D  E  of  the  plan. 
Upon  any  convenient  straight  line  lay  off 
D'  A*  in  Fig.  420,  in  length  equal  to  D' 
A",  and  from  A4  set  off,  at  right  angles  to 
it,  A4  C3,  in  length  equal  to  E  C1  of  the 
plan.  Then  D4  A*  C3  shows  in  the  flat  one- 
half  of  the  diagonal  face  of  the  roof,  or 
what  is  represented  by  DEC1  in  the  plan. 
At  right  angles  to  D4  C'  draw  the  remain- 
ing portion  of  the  stay  not  used  in  con- 
nection with  the  true  face,  placing  it  in 
such  a  manner  that  the  point  O3,  corre- 
sponding to  0  of  the  hip  section,  shall  fall 
upon  the  line  D4  C3,  which  represents  the 
angle  of  the  hip.  Through  the  point  8  of 
the  section  L"  M7,  corresponding  to  8  of  the 
section  L"  M3  of  Fig.  418,  draw  a  line 
parallel  to  D4  C',  as  shown  by  S'  Y'.  Then 
S'  Y'  corresponds  to  S  Y  of  the  true  face 
in  Fig.  418. 

Space  the  profile  L°  M7  into  the  same 
parts  as  used  in  laying  off  the  stretchout 
W  V,  and  through  the  points  draw  lines 
parallel  to  D'  G\  cutting  the  line  S"  A4, 
which,  being  the  center  line  of  the  octag- 
onal side  of  the  roof,  is  also  the  miter 
line  between  the  two  arms  of  the  hip  mold- 
ing. From  the  points  of  intersection 
in  the  line  D4  A4,  at  right  angles  to  S3 
Y1,  draw  lines  cutting  S1  Y1,  giving  the 
points  marked  8,  9,  10,  11,  12,  13  and  14.  For  con- 
venience in  using  one  stretchout  for  the  entire  pat- 
tern, transfer  these  points  to  the  line  S  Y  of  the  true 
face  in  Fig.  418,  from  which,  at  right  angles  to  S  Y, 
draw  lines  cutting  the  corresponding  measuring  lines 
of  the  stretchout.  Then  a  line  traced  through  these 


For  the  portion  of  the  pattern  corresponding  to      points  of  intersection,  as  shown  from  S'  to  X,  will  com- 
the  part  of  the  profile  which  miters  against  the  other  [  plete  the  pattern. 


236 


Tlie  New  Metal   Worker  Pattern  Book. 


PROBLEM   no. 


Patterns  for  the  Fascias  of  a  Hip  Molding  Finishing  a  Curved  Mansard  Roof  which  is  Square  at  the 

Base  and  Octagonal  at  the  Top. 


The  conditions  involved  in  this  problem  do  not 
differ  greatly  from  those  given  in  Problems  SO,  81  and 
82,  near  which  it  should  properly  be  classed.  In  this 
case,  however,  the  profile  of  an  entire  roof  is  under 
consideration  instead  of  that  of  a  simple  molding  or 
vase,  but  the  problem  is  here  introduced  as  being 
closely  related  in  feature  to  several  of  the  foregoing 
problems. 

C  D  E  F  of  Fig.  421  represents  the  plan  of  the 
roof  at  its  base,  while  V  G  II  W  represents  the  plan  at 
the  top.  It  will  be  seen  that  the  roof  is  nearly  square 
at  the  foot  of  the  rafters  and  octagonal  at  the  top.  The 
same  conditions  may  arise  where  the  corners  of  the 
roof  are  chamfered,  starting  at  nothing  at  the  bottom 
and  increasing  to  a  considerable  space  at  the  top,  with- 
out reference  to  forming  an  octagon.  D  G  II  E  in  the, 
plan  represents  a  chamfer  or  transition  piece  in  the  con- 
struction of  a  roof  which,  as  above  described,  is  square 
at  the  base  and  octagonal  at  the  top.  This  part  is  rep- 
resented in  elevation  by  D3  G"  II*  E3.  The  elevation  is 
introduced  here  not  for  any  use  in  pattern  cutting,  but 
simply  to  show  the  relation  of  parts.  In  the  sectional 
view  of  the  roof  0  A  B  the  outer  line  0  B  represents 
the  surface  of  the  fascias  of  which  the  patterns  are  re- 
quired, the  inner  curve  showing  the  line  of  the  roof 
boards  and  the  depth  of  the  sink  strips.  As  it  is  in  the 
plan  that  the  miter  lines  are  shown  it  will  be  necessary 
to  develop  the  pattern  from  the  plan.  Assuming  then 
that  one  of  the  square  sides,  as  E  F  W  II,  is  to  be 
done  first,  it  will  be  necessary  to  place  a  profile  so  that 
its  projection  O  A  shall  lie  across  this  part  of  the  roof, 
al".  as  shown  by  0'  A1  B1. 

Divide  the  profile  O1  B1  into  any  convenient  num- 
ber of  equal  spaces,  and  from  the  points  of  division 
drop  lines  parallel  to  E  F,  the  side  of  the  roof,  cutting 
the  miter  line  E  II.  Upon  any  line  at  right  angles  to 
this  side  of  the  roof,  as  O2  B*,  lay  off  a  stretchout 
through  the  points,  in  which  draw  the  usual  measuring 
lines.  Cut  these  measuring  lines  bylines  drawn  verti- 
cally from  the  points  in  E  H.  Then  a  line  traced 
through  these  points  of  intersection,  as  shown  by  E3 
H',  will  be  the  line  of  the  pattern  corresponding  to  the 


line  E  H  in  the  plan.  The  width  of  the  flange  or  fascia 
forming  the  hip  finish  may  be  obtained  as  described  in 
Problem  6,  and  the  corner  piece  drawn  in  to  agree  with 
the  original  design,  as  shown  by  S'  T' 

If  it  is  desirable  to  produce  an  elevation  of  this 
angle  of  the  roof  it  can  be  done  bv  dividing  the  profile 
0  B  by  the  same  points  as  were  used  in  dividing  O1  B', 
from  which  horizontal  lines  can  be  drawn  to  the  left 
intersecting  with  the  lines  of  corresponding  number 
previously  erected  from  the  miter  line  E  II.  A  line, 
E3  II3,  drawn  through  >he  points  of  intersection  will 
with  D*  G'  give  the  correct  elevation  of  the  transition 
side. 

For  the  pattern  of  this  side  it  will  be  necessary  to 
first  construct  a  section  upon  its  center  line,  P  R  of  the 
plan.  At  any  convenient  place  outside  of  the  plan 
draw  a  duplicate  of  P  R  parallel  to  it,  as  shown  by 
P1  A',  and  from  the  point  A1  erect  a  perpendicular, 
A1  B',  in  length  equal  to  A  B  of  the  original  section. 
In  A'  B'  set  off  points  corresponding  to  the  points  in 
A  B,  and  through  them  draw  horizontal  lines,  as  shown. 
Place  the  "{-square  parallel  to  A'  B1,  and,  bringing  it 
against  the  points  in  E  II  previously  obtained  from  the 
profile  0'  B1,  cut  corresponding  measuring  lines.  Then 
a  line  traced  through  these  points  of  intersection,  as 
shown  by  B'  P',  will  complete  the  diagonal  section 
corresponding  to  P  R  in  the  plan.  From  this  diagonal 
section  take  a  stretchout,  which  lay  off  on  the  straight 
line  corresponding  to  P  R  produced,  all  as  shown  by 
P3  B3.  Through  the  points  in  P"  B3  draw  the  usual 
measuring  lines.  With  the  T-square  placed  parallel  to 
this  stretchout  line,  and  brought  successively  against 
the  points  in  R  II,  cut  the  measuring  lines,  as  shown. 
Then  a  line  traced  through  these  points  of  intersection, 
as  shown  by  E1  to  II',  will  be  one  side  of  the  required 
pattern.  In  like  manner,  having  transferred  points 
from  E  II  across  to  the  corresponding  line  D  G,  cut  the 
measuring  lines  from  it,  which  will  give  the  other  side 
of  the  required  pattern.  The  width  of  fascias  (whose 
intersection  forms  a  panel  in  this  case)  may  be  obtained 
as  suggested  above  and  as  given  in  Problem  6. 

In  locating  the  points  N1  and  M1  of  this  pattern  it 


Pattern  Probkras, 


227 


IV 

_—-  _ "nx/*- 


• / u  i a 


:.— Bitterns  /or  the  Fascias  o/  a  Ht_p  Molding  Finishing  a  Curved  Mansard  Roof  Which  is  Square  at  the  Eaves  and 

Octagonal  at  the  Top. 


238 


Tlte  Xew  Metal    Worker  Pattern  Bwk. 


is  desirable  for  the  sake  of  design  tliat  they  be,  when 
finished  and  in  position,  at  the  same  vertical  distance 
below  the  corniee  as  are  the  points  S  and  T  on  the 
square  sides  of  the  roof.  To  accomplish  this  it  will  be 
necessary  to  go  back  to  the  points  S'  and  T1,  in  the  first 
pattern  obtained,  and  from  them  carry  lines  back  into 
the  stretchout  line  O*  B4,  where  they  are  numbered 
10£  and  11£.  Their  positions  may  now  be  transferred 
by  means  of  the  dividers  to  the  normal  profile  O  B, 
where  their  vertical  bights  can  be  measured  on  the  line 
A  B,  as  shown,  and  transferred  again  to  the  vertical 
line  A1  B1  of  the  diagonal  section.  It  is  only  neces- 
sary now  to  carry  them  across,  as  shown,  to  the  profile 
P'  B',  where  their  distances  from  adjacent  points  may  be 
measured  by  the  dividers  and  placed  upon  the  stretch- 
out line  P2  B'.  By  similar  means  the  appearance  of 
this  panel  both  in  the  plan  and  in  the  elevation  may  be 


completed  if  s<>  desired,  all  of  \vhichwillbemade  clear 
by  inspection  of  the  drawing. 

In  the  ease  of  very  large  roofs,  where  the  develop- 
ment of  a  profile  or  a  pattern  to  the  full  size  would  be 
impracticable,  it  is  possible  to  perform  the  work  to  a 
scale  of  li  or  :5  inches  to  the  foot;  after  which  full 
size  patterns  of  parts  of  convenient  size  may  be  ob- 
tained by  multiplying  their  various  dimensions  by  8 
or  4. 

As  the  patterns  for  the  roll,  usually  finishing 
the  hip,  are  properly  included  under  the  head  of 
Flaring  Work,  which  subject  is  treated  in  the  fol- 
lowing section  of  this  chapter,  they  will  not  be 
given  here.  The  radii  from  which  they  can  be  ob- 
tained, however,  may  be  derived  from  the  diagonal 
section  in  the  manner  described  in  the  following 
problem. 


PROBLEM    in. 


To  Obtain  the  Curves  for  a  Molding  Covering  the   Hip  of  a  Curved  Mansard  Roof. 


The  method  of  obtaining  the  pattern  of  the  fascias 
of  a  molding  covering  a  curved  hip  has  been  given  in 
Problem  G.  As  it  is  necessary  in  obtaining  the  pat- 
terns of  the  molded  portion  or  roll,  that  the  curve  of  the 
hip  should  be  established,  this  problem  really  consists 
of  developing  from  the  normal  profile  of  the  roof  a 
profile  through  the  hip,  or,  in  other  words,  a  diagonal 
section  of  the  mansard. 

Let  A  E  B  iii  Fig.  422  represent  the  plan  of  a 
mansard  roof  or  tower,  the  elevation  of  which  is  shown 
by  II  K,  over  the  hip  of  which  a  molding  of  any  given 
profile  is  to  be  fitted,  in  this  case  a  three-quarter  bead, 
the  diagonal  line  E  F  in  the  plan  representing  the 
angle  of  the  hip  as  it  would  appear  if  viewed  from  the 
top.  At  any  convenient  point  parallel  to  E  F,  and 
equal  to  it,  drawK'  F',  and  from  F'  erect  a  perpendicu- 
lar, F1  K1,  in  length  equal  to  the  vertical  line  in  eleva- 
tion G  K.  Divide  G  K  and  F'  K'  into  the  same  num- 
ber of  equal  spaces.  From  the  points  in  G  K  draw 
lines  cutting  the  profile  H  K,  as  shown,  and  from  the 
points  thus  obtained  in  H  K  drop  lines  vertically,  pro- 
ducing them  until  they  cut  the  diagonal  line  E  F  of  the 
plan,  as  shown.  Through  the  points  in  F1  K'  draw 


measuring  lines  in  the  usual  manner,  and  intersect  them 
by  lines  erected  perpendicularly  to  E  F  from  the  points 
therein.  Then  a  line  traced  through  these  points  of 
intersection,  as  shown  by  E1  K',  will  be  the  profile  to 
which  the  molding  covering  the  hip  is  to  be  raised. 

Inasmuch  as  in  the  usual  process  of  mold  raising 
all  curves  must  be  considered  as  segments  of  circles, 
to  accommodate  both  the  adjustment  of  the  machine 
used  and  the  describing  of  the  patterns,  the  curved 
line  E1  K'  just  obtained  must  be  so  divided  that  each 
section  or  segment  will  approach  as  nearly  as  possible 
an  arc  of  a  circle.  In  this  case  the  section  from  E1  to 
L  will  be  found  to  correspond  to  an  arc  struck  from 
a  center,  M,  while  the  section  from  L  to  K'  corresponds 
to  an  arc  struck  from  a  center  not  shown  in  the  engrav- 
ing, but  which  will  be  found  by  the  intersection  of  the 
lines  L  N  and  K1  N1  produced. 

In  the  lower  part  of  Fig.  423  is  shown  an  enlarged 
section  of  the  hip  molding,  including  the  fascias,  as  it 
would  appear  at  the  bottom  of  the  hip,  and  above  it 
another  section  taken  at  the  top,  which  has  been  de- 
rived from  the  normal  section  or  section  at  the  bottom 
by  the  method  used  and  explained  in  Problems  105, 


I'n Htm   Prvbkms. 


lot!    and     K>7,    previously    demonstrated.     A    dotted      the  roll  require  trimming  after  being  raised  so  that  the 

reproduction    of    the    lines    oi    the    upper    suction    is   j   roll  may  have  an  equal  projection  throughout  its  course. 


SE.CTIONATTOP 


Fig.  423.— Enlarged  Sections  Through  Hip  Finish  at 
Top  and  Bottom,  Showing  Change  in  Flare  of  Fascias. 


Fig.  42%.— Diagonal  Section  of  a  Curved  Mansard  Roof  Obtained  for  the  Purpose  of  Mold  Raising. 


placed  here  to  show  the  change  in  the  flare  that  takes 
place  between  fascias  in  going  from  the  bottom  to  the 
top  of  the  hip,  thus  showing  that  the  outer  edges'  of 


Methods  of  obtaining  the  patterns  of  curved  mold- 
ings will  be  found  in  the  following  section  of  this 
chapter. 


240 


The  New  Metal    Worker  l*uttcru   JJuuk. 

SECTION  2. 


(FLARING  WORK.) 


It  will  be  well  to  place  before  the  reader  here  a 
clear  statement  of  the  class  of  problems  he  may  ex- 
pect to  meet  with  under  this  head.  It  will  include 
only  the  envelopes  of  such  solid  figures  as  have  for  a 
base  the  circle,  or  any  figure  of  equal  or  unequal  sides 
which  may  be  inscribed  within  a  circle,  and  which 
terminate  in  an  apex  located  directly  over  the  center 
of  the  base. 

According  to  the  definition  of  an  inscribed  poly- 
gon (l)ef.  66),  its  angles  must  all  lie  in  the  circum- 
ference of  the  same  circle.  So  the  angles  or  hips  of  a 
pyramid  whose  base  can  be  inscribed  in  a  circle  must 
lie  in  the  surface  of  a  cone  whose  base  circumscribes 
its  base  and  whose  altitude  is  equal  to  that  of  the 
pyramid.  Therefore  the  circle  which  describes  the 
pattern  of  the  base  of  the  envelope  of  such  a  cone  will 
also  circumscribe  the  pattern  of  the  base  of  the  pyra- 
mid contained  within  it.  The  envelopes  of  such  solids, 
therefore,  as  scalene  cones,  scalene  pyramids  and 
pyramids  whose  bases  cannot  be  inscribed  within  a 
circle  are  not  adapted  to  treatment  by  the  methods 
employed  in  this  section.  Even  the  envelope  of  an 
elliptical  cone  cannot  be  included  with  this  class  of 
problems  because  it  possesses  no  circular  section  upon 
which  its  circumference  at  any  fixed  distance  from 
the  apex  can  be  measured. 


In  this  connection  it  is  proper  to  call  attention  to 
the  difference  between  a  scalene  cone  and  a  right  cone 
whose  base  is  oblique  to  its  axis.  According  to  Ih'li- 
nition  96,  a  scalene  cone  is  one  whose  axis  is  inclined 
to  the  plane  of  its  base,  and  according  to  Definition  !M 
the  base  of  a  cone  is  a  circle.  As  any  section  of  a 
cone  taken  parallel  to  its  base  is  the  same  shape  as  its 
base,  any  section  of  a  scalene  cone  taken  parallel  to 
its  base  must  be  a  circle,  and  any  section  taken  at 
right  angles  to  its  axis  could  not,  therefore,  be  a  circle, 
but  would  be  elliptical.  Again,  as  any  section  of  a 
right  cone  (Def.  95)  at  right  angles  to  its  axis  is  a 
circle,  if  its  base  be  cut  off  obliquely,  such  base  would. 
according  to  Definition  113,  be  an  ellipse.  There- 
fore, since  its  horizontal  section  is  a  circle,  its 
envelope  may  be  obtained  by  methods  employed  in 
this  section.  (See  Problem  136.)  And  since  the  sec- 
tion of  a  scalene  cone  taken  at  right  angles  to  its  axis 
is  an  ellipse,  the  scalene  cone  becomes  virtually  an 
elliptical  cone  with  an  oblique  base — that  is,  with  a  base 
cutoff  at  such  an  angle  as  to  produce  a  circle — and,  as 
stated  above,  cannot  be  included  in  this  section. 

The  principles  governing  the  problems  of  this 
section  are  given  in  Chapter  V,  beginning  on  page  79, 
which  the  reader  will  find  a  great  help  in  explaining 
anything  which  he  may  fail  to  understand. 


PROBLEM    112. 
The  Envelope  of  a  Triangular  Pyramid. 

Let  A  B  C  of  Fig.  424   be    the    elevation  of    the      the  point  K  erect  K  H,  perpendicular  to  F  K  and  equal 
pyramid,  and  E  F  Gr  of  Fig.  425  the  plan.     From  the      in  length  to  the  hight  of  the  pyfamid,  as  shown   by 


Fig.  424. -Elevation. 


Fig.  425.— Plan. 
The  Envelope  of  a  Trianr/ular  Pyramid. 


Fig.  426.-Pattern. 


center  K  draw  the  lines  E  K,  F  K  and  G  K  in  the  plan, 
representing  the  angles  or  hips  of  the  pyramid.     From 


A    1)  of   the  elevation.      Draw  the  hypothenuse  F  II, 
which  then  represents  the  length  of  the  corner  lines. 


Pattern   Problems. 


241 


From  anv  point,  as  L  of  Fig.  426,  for  center,  with 
radius  equal  in  V  II,  describe  the  arc  M  N  O  I  indef- 
inite! v,  and  draw  L  M.  From  M  set  off  the  chord  M 
N,  in  length  equal  to  the  sideF  G  of  the  plan.  lu  like 


manner  set  off  N  0  and  0  I  respectively,  equal  to  G  E 
and  E  F  of  the  plan.  Connect  I  and  L,  as  shown, 
and  draw  L  ()  and  L  N.  Then  L  I  0  N  M  is  the 
pattern  sought. 


PROBLEM     113. 

The  Envelope  of  a  Square  Pyramid. 


Let  E  A  C  of  Fig.  427  be  the  elevation  of  the 
pyramid,  and  F  II  K  L  of  Fig.  428  the  plan.  The 
diagonal  lines  F  K  and  L  II  represent  the  plan  of  the 
angles  or  hips,  and  G  a  point  corresponding  to  the 
apex  A  of  the  elevation.  From  the  apex  A  drop  the 


or  circumscribe  the  pattern,  as  shown  in  the  diagram 
From  any  center,  as  M,  Fig.  429,  with  a  radius  equal 
to  A  D,  describe  an  arc,  as  P  R  0  S  N,  indefinitely 
and  draw  M  P.      From  P,  on  the  arc  drawn,     set  off 
a  chord,  P  R,  in  length  equal  to  one  of  the  sides  of 


_     N 


Fig.  437.— Elevation. 


Fig.  428.-Plan. 
The  Envelope  of  a  Square  Pyramid. 


Fig.  429.— Pattern. 


line  A  B  perpendicular  to  the  base  E  C.  Prolong  E 
0  in  the  direction  of  D,  making  BD  equal  to  G  F,  one 
of  the  angles  of  the  plan.  Connect  D  and  A.  Then 
A  D  wiir  be  the  slant  hight  of  the  article  on  one  of  the 
corners,  and  the  radius  of  an  arc  which  will  contain 


the  pyramid  shown  in  the  plan.  From  R  set  off 
another  chord,  R  O,  in  like  manner,  and  repeat  the 
same  operation,  obtaining  0  S  and  S  N.  Draw  the  lines 
M  N,  M  S,  M  O  and  M  R.  Then  M  N  S  0  R  P  will 
be  the  required  pattern. 


PROBLEM    114. 
The  Envelope  of  a  Hexagonal  Pyramid. 


Let  H  G  I  of  Fig.  430  represent  the  elevation  of 
a  hexagonal  pyramid,  of  which  D  F  C  L  B  E  of  Fig. 
431  is  the  plan.  The  first  step  is  to  construct  a  section 
on  a  line  drawn  from  the  center  of  the  figure  through 
oue  of  its  angles  in  the  plan,  as  A  B.  From  the  center 
A  erect  A  X  perpendicular  to  A  B,  making  it  equal  to 


the  straight  hight  of  the  article,  as  shown  in  the  eleva- 
tion by  G  K.  Draw  the  hypothenuse  B  X.  Then  X 
represents  the  apex  and  XB  the  side  of  a  right  cone,  the 
plan  of  the  base  of  which,  if  drawn,  would  circum- 
scribe the  plan  of  the  hexagonal  pyramid.  From  any 
convenient  center,  as  X1  of  Fig.  432,  with  X  B  of 


242 


Tlie  Xew  Metal    Worker  Pattern   Book. 


Fig.  431  as  radius,  describe  an  arc  indefinitely,  as 
shown  by  the  dotted  line.  Through  one  extremity  of 
the  arc  to  the  center  draw  a  line,  as  shown  by  D'  X1. 


B'  L'  in  the  arc  thus  obtained  draw  lines  to  the  center, 
as  shown  by  E'  X',  B1  X1,  etc.,  which  will  represent 
the  angles  of  the  completed  shape,  and  serve  to  locate 


Fig.  430.— Elevation. 


Fig.  431.-Plan. 

The  Envelope  of  a  Hexagonal  Pyramid. 


Fig.  432.-Pattern. 


With  the  dividers  set  to  a  space  equal  to  any  side  of  the 
plan,  as  D  E,  commencing  at  D1,  set  off  this  distance  on 


the  bends  to  be   made   in   process    of    forming    up. 
Then  X1   D1   E1   B1   L1  C'  F1  D3  will  be  the  complete 


the  arc  six  times,  as  shown.     From  the  several  points  E'  ;  pattern. 

PROBLEM   115. 

The  Envelope  of  the  Frustum  of  a  Square  Pyramid. 

In  Fig.  433,  let  G  H  K  I  be  the  elevation  of  the      Construct  a  diagonal  section  on  the  lin?  A  P  as  fol- 
article,  0  A  E  D  the  plan  of  the  larger  end  and  L  M      lows :     Erect  the  perpendicular  P  F,  making  it  equal 


Fig.  433.-Plan  and  Elevation. 


Fig.  434.— Pattern. 
The  Envelope  of  the  Frustum  of  a  Square  Pyramid. 


0  N  the  plan  of  the  smaller  end.     Produce  the  hip 
lines  C  L,  A  M,  etc.,   in  the   plan   to  the  center  P. 


to  the  straight  hight  of  the  article,  as  shown  by  R  K 
of   the   elevation.     Likewise  erect   the  perpendicular 


Pattern  Problems. 


243 


M  B  of  the  same  length.  Draw  F  B  and  A  B.  Then 
P  A  B  F  is  the  diagonal  section  of  the  article  upon 
the  line  P  A.  Produce  A  B  indefinitely  in  the  direc- 
tion of  X,  and  also  produce  P  F  until  it  meets  A  B 
extended  in  the  point  X.  Then  X  is  the  apex  of  a 
right  cone  and  X  A  the  side  of  the  same,  the  base  of 
which,  if  drawn,  would  circumscribe  the  plan  C  A  E  D. 
Therefore,  from  any  convenient  center,  as  X'  of  Fig. 
434,  with  X  A  as  radius,  describe  the  arc  C'  D'  E'  A1 
C",  and  from  the  same  center,  with  radius  X  B,  draw 


the  arc  L1  N' 0' M' L1,  both  indefinitely.  Draw  C' 
X',  cutting  the  smaller  arc  in  the  point  L1.  Make  the 
chord  C'  D1  equal  in  length  to  one  side,  <J  D,  of  the 
plan,  and  D'  E1  to  another  side,  D  E,  of  the  plan, 
and  so  on,  until  the  four  sides  of  the  base  have  been 
set  off.  Draw  D'  X1,  E'  X',  etc.,  cutting  the  arc 
L1  LJ  in  the  points  N1,  O1,  etc.  Then  D1  N1,  E1  O1 
and  A1  M'  will  represent  the  lines  of  the  bends  in 
forming  up  the  pattern.  Draw  the  chords  L'  N',  N1  O1, 
etc.,  thus  completing  the  pattern. 


PROBLEM   116. 


The  Envelope  of  the  Frustum  of  an  Octagonal  Pyramid. 


<•• 


Tier.  437.-Pattern. 
The  Envelope  of  the  Frustum  of  an  Octagonal  Pyramid. 


Fig.  435  shows  the  elevation  and  Fig.  436  the 
plan  of  the  frustum  of  an  octagonal  pyramid.  The 
•  first  step  in  developing  the  pattern  is  to  construct  a 
diagonal  section,  the  base  of  which  shall  correspond  to 
one  of  the  lines  drawn  from  the  center  of  the  plan 
through  one  of  the  angles  of  the  figure,  as  shown  by 
G.  B.  Erect  the  perpendicular  G  C  equal  to  the 
straight  hight  of  the  frustum,  as  shown  by  N  M  of  the 
elevation,  and  at  b  erect  a  perpendicular,  b  A,  of  like 
length.  Draw  B  A  and  A  C.  Then  G  B  A  C  is  a 
section  of  the  article  as  it  would  appear  if  cut  on  the 
line  G  B.  Produce  B  A  indefinitely  in  the  direction 
of  X,  and  likewise  prolong  G  C  until  it  intersects  B  A 
produced  in  X.  Then  X  is  the  apex  and  X  B  the 
side  of  a  right  cone,  the  plan  of  which,  if  drawn, 
would  circumscribe  the  base  of  the  frustum.  From 
any  convenient  center,  as  X',  Fig.  437,  with  radius 
X  B,  describe  an  arc  indefinitely,  as  shown  by  the 
dotted  line  E1  E'  of  the  pattern,  and  from  the  same 
center,  with  X  A  for  radius,  describe  the  arc  e'  e'  of 
the  pattern.  Through  one  extremity  of  the  arc  E1  Ei 
to  the  center  draw  a  straight  line,  as  shown  by  E'  X'^ 
cutting  the  smaller  arc  in  the  point  e1.  Set  off  on  the 
arc  E1  Es  spaces  equal  to  the  sides  of  the  plan  of  the 
base  of  the  article  and  connect  the  points  by  chords. 
Thus  make  E1  P'  of  the  pattern  equal  to  E  P  of  the 
plan,  and  so  on.  Also  from  these  points  in  the  arc 
draw  lines  to  the  center,  cutting  the  arc  e1  e',  as  shown. 
Connect  the  points  thus  obtained  in  this  arc  by  chords, 
as  shown  by  e1  p\  p'  d\  dl  o1,  etc.  Then  e'  E1  E"  e'  will 
be  the  pattern  sought. 


TAe  New  Metal  Worker  Pattern  Hook. 

PROBLEM   117 
The  Envelope   of  the  Frustum  of  an  Octagonal  Pyramid  Having  Alternate  Long  and  Short  Sides. 


In  Fig.  438,  let  I  M  B  N  O  P  K  L  be  the  phm  of 
the  article  of  which  G  H  F  E  is  the  elevation.  The 
first  thing  to  do  in  describing  the  pattern  is  to  construct 
a  section  corresponding  to  a  line  drawn  from  the  center 


Produce  S  K  and  B  A  until  they  meet  in  the  point  X. 
Then  X  is  the  apex  and  X  B  is  the  side  of  a  cone,  the 
base  of  which,  if  drawn,  would  circumscribe  the  plan 
of  the  article.  From  any  convenient  center,  as  X', 
Fig.  439,  with  radius  equal  to  X  B,  describe  an  arc. 
as  shown  by  M1  M".  Draw  X'  M'  as  one  side  of  the 
pattern.  Then,  starting  from  M1,  set  off  chords  to  the 
arc,  as  shown  by  M1  B',  B' N1,  etc. ,  equal  to  and  corre- 
sponding with  the  several  sides  of  the  article,  as  shown 
by  M  B,  B  N,  etc.,  in  the  plan.  From  these  points. 
B1,  N',  etc.,  in  the  arc,  draw  lines  to  the  center  X'. 


Fig.  438.— Plan  and  Elevation. 


Fig.  439.-Pattern. 


The  Envelope  of  the  Frustum  of  an  Octagonal  Pyramid  Having  Alternate  Long  and  Short  Sides. 


to  one  of  the  angles  in  the  plan,  as  SB.  At  S  erect 
'  the  perpendicular  S  B,  in  length  equal  to  the  straight 
hight  of  the  article,  as  shown  by  C  D  of  the  elevation. 
Upon  the  point  b  erect  a  corresponding  perpendicular, 
as  shown  by  b  A.  Draw  R  A  and  A  B.  Then  B  A 
B  S  is  a  section  of  the  article  taken  upon  the  line  S  B. 


From  X1,  with  X  A  as  radius,  describe  an  arc  cutting 
these  lines,  as  shown  by  m'  m*.  Connect  the  points  of 
intersection  by  straight  lines,  as  shown  by »«.'  //',  b'  n\  //'  </. 
etc.  Then  in'  nr  M2  M1  will  be  the  pattern  sought. 
and  the  lines  B1  i1,  N1  »',  etc.,  will  represent  the  lines 
of  bends  to  be  made  in  forming  up  the  article. 


PROBLEM    118. 

The  Pattern  of  a  Square  Spire  Mitering  Upon  Four  Gables. 


In  Fig.  440,  let  B  F  H  C  be  the  elevation  of  a 
square  spire  which  is  required  to  miter  over  four  equal 
gables  in  a  pinnacle,  the  plan  of  which  is  also  square. 


Produce  F  B  and  H  C  until  they  meet  in  A,  which 
will  be  the  apex  of  the  pyramid  of  which  the  spire  is  a 
section.  Draw  the  axis  A  G,  and  at  right  angles  to  it, 


Pattern  Problems. 


245 


from  the  lowest  point  of  contact  between  the  spire  and 
the  gable,  as  F,  draw  F  G.  Then  F  (1  will  represent 
the  half  width  of  one  of  the  sides  of  the  pyrarnicl  at  the 
base,  and  A  F  will  represent  the  length  of  a  side 
through  its  center.  From  any  convenient  point,  as  A1 


spaces  of  the  extent  of  G'  G3,  as  shown  by  G"  g,  gg1  and  g' 
O.  Draw  g1  A1,  g  A1  and  O  A'.  Make  A1  B1  equal  to  A 
13  of  the  elevation,  and  through  B1  draw  a  perpendicu- 
lar to  A'  F1,  as  shown.  Draw  lines  corresponding  to  it 
through  the  other  sections  of  the  pattern.  Make  A1  D1 


Fig.  440.— Elevation  of  Spire. 


—-..A1 

' 


Fig.  441.— Pattern. 


The  Pattern  of  a  Square  Spire  Mitering  Upon  Four  Gables. 


in  Fig.  441,  draw  A1  F1,  in  length  equal  to  A  F.  From 
F'  set  off,  perpendicular  to  A'  F1,  on  each  side  a  space 
equal  to  F  G  of  the  elevation,  as  shown  by  F1  G'  and 
F'  (¥.  From  G'  and  G"  draw  lines  to  A1,  as  shown. 
From  A1  as  center,  and  with  A1  G2  as  radius,  describe 
an  arc,  as  shown  by  G3  0,  in  length  equal  to  three 


equal  to  A  D,  and  draw  D1  G1  and  D1  G5.  Set  the 
compasses  to  G2  D',  and  from  G"  and  y  as  centers 
describe  arcs  intersecting  at  d.  Draw  d  g  and  d  G5,  as 
shown.  Eepeat  the  same  operation  in  the  other 
sections  of  the  pattern,  thus  completing  the  required 
shape. 


PROBLEM    119. 

The  Pattern  of  an  Octagon  Spire  Mitering  Upon  Eight  Gables. 


Let  A  G  L  in  Fig.  442  be  the  elevation  of  the 
spire,  and  M  O  P  the  half  plan.  From  the  point  G, 
which  represents  the  lowest  point  of  the  angle  or  valley 
between  the  gables,  to  H,  which  represents  the  meet- 


ing of  the  valleys  and  ridges  at  T  in  the  plan,  draw 
the  line  G  H,  cutting  the  side  A  C  extended  in  the 
point  D.  Draw  any  line,  as  A'  D1  in  Fig.  443,  upon 
which  to  construct  the  pattern.  Make  A1  C'  equal  to 


Tlte  New  Metal    Worker   Pattern    I  took. 


\  ( '  of  the  elevation,  and  A'  D'    equal   to   A  D  of  the 
elevation.     Through  D'  draw  the  horizontal  line  1  0,  as 


Draw  A'  it  and  A'  1.  Set  the  dividers  to  A'  1  as 
radius,  and  from  A'  as  center  describe  the  arc  1  8  in- 
definitely. Set  the  dividers  to  1  0,  and  step  off  as 
many  spaces  on  the  arc  as  there  are  sides  in  the  spire. 
Draw  the  lines  A1  2,  A1  3,  etc.,  to  A1  S,  which  represent 
the  angles  of  the  spire  and  the  bends  in  the  pattern. 
Draw  C'  0  and  C'  1  in  the  first  section  of  the  pattern. 
Set  the  dividers  to  C;  1,  and  from  1  and  2  as  centers 
describe  intersecting  arcs,  as  shown  by  C*.  In  like 
manner  describe  similar  intersecting  arcs  at  the  points 


D'        0 


Fig.  442.— Plan  unil  Klevation  of  Spire. 


Tig.  443.— Pattern. 


The,  Pattern  of  an  Octagon  Spire  Mitering  Upon  Eight  Gables. 


shown.     From  D1  set  off  D'  0,  equal  to  E  F  of  the  ele- 
vation, and  likewise  set  off  D'  1,  of  the  same  length. 


C3,  C4,  etc.     Draw  lines  from  these  points  to  the  points 
1,  2,  3,  4,  etc.,  as  shown,  thus  completing  the  pattern. 


PROBLEM    120. 

The  Pattern  of  an  Octagon  Spire  Mitering  Upon  Four  Gables. 


In  Fig.  444-,  let  B  E  Z  U  be  the  elevation  of  an  octa- 
gon spire,  mitering  down  upon  four  gables  occurring 
upon  a  square  pinnacle.  Continue  the  side  lines  until 
they  intersect  in  the  apex  A.  Draw  the  center  line  A 
H,  and  from  the  point  Gdraw  G  H  perpendicular  to  the 
center  line,  showing  half  the  width  of  one  of  the 
sides  at  the  point  G.  Ry  inspection  of  the  elevation  it 


will  be  seen  that  one-half  the  sides  will  be  notched  at 
the  bottom  to  fit  over  the  gables,  while  the  others  will 
be  pointed  to  reach  down  into  the  angles  or  valleys  be- 
tween the  gables. 

To  ascertain  the  correct  length  upon  the  center 
line  of  one  of  the  pointed  sides  it  will  be  necessary  to 
construct  a  section  through  one  of  the  valleys,  for  in- 


/  '<  i  Ili-rn  Problems. 


241 


stance,  upon  tlie  line  M'  N'  of  the  plan.  Through  the 
point  J  of  the  elevation  draw  the  line  .)  M  at  right 
angles  to  the  center  line,  extending  it  to  the  left  in- 
definitely, and  from  the  point  M  set  off  upon  this  line 
the  distance  M  N,  equal  to  M'  N1  of  the  plan.  Draw 


equal  to  A  E  of  the  elevation,  etc.  Through  E'  draw 
a  perpendicular  equal  in  length  to  the  width  of  a  side 
at  the  point  E,  or  to  twice  G  H,  as  shown  in  the  ele- 
vation, placing  one-half  on  each  side  from  E1,  all  as 
shown  by  L  K.  From  L  and  K  draw  lines  to  A1.  From 
A'  as  center,  with  A1  L  as  radius,  describe  an  arc,  as 
shown  by  L  U,  indefinite  in  length.  Set  the  dividers 
to  the  space  L  K,  and  step  off  spaces  from  L,  as  L  Y, 
Y  X,  etc.,  until  as  many  sides  are  set  off  as  are  required 
in  the  spire — in  this  case  eight.  Draw  the  lines  A1  Y, 
A1  X,  etc.  From  the  point  D',  which,  as  will  be  seen 
by  D  in  the  elevation,  corresponds  to  the  top  of  the 
gable,  draw  lines  to  the  points  L  and  K,  which  gives 
the  pattern  for  the  notch  in  the  first  section.  Set 
the  dividers  to  L  D1  as  radius,  and  from  X  and  Y  as 
centers  describe  arcs  intersecting  at  W.  Draw  W 
X  and  "W  Y,  and  repeat  this  upon  all  the  alternate 
sides  throughout  the  pattern,  as  shown,  locating  the 


Fiir.  444.-Plan  and  Elevation  of  Spire.  Fig;.  445.— Pattern. 

The  Pattern  of  an  Octagon  Spire  Mitering  Upon  Four  Gables. 


N  P,  and  extend  the  side  A  E  until  it  intersects  this  line 
at  F.  Then  A  F  will  be  the  correct  length  through 
the  center  line  of  one  of  the  long  sides. 

To  describe  the  pattern  first  draw  A1  F1  in  Fig. 
445,  equal  to  A  F  of  the  elevation,  and  set  off  points 
on  it  corresponding  to  points  in  A  F.  Thus  make 
A1  B'  equal  to  A  B,  A1  D1  equal  to  A  D,  and  A'  E1 


points  0  and  P.  For  the  pattern  of  the  point,  take 
a  space  between  the  points  of  the  dividers  equal 
to  L  F',  and  from  L  and  Y  as  centers  describe  small 
arcs  intersecting. at  M,  and  draw  M  L  and  M  Y.  With 
the  same  radius  repeat  the  operation  upon  the  inter- 
mediate sides,  establishing  the  points  V,  H  and  I,  thus 
completing  the  pattern. 


The  New  Metal   Worker  Pattern  Book. 

PROBLEM    121. 


Pattern  for  an  Octagon  Spire  Mitering  upon  a  Roof  at  the  Junction  of  the  Ridge  and  Hips. 


In  Fig.  446,  let  ABC  represent  the  front  eleva-   i  vation    let  W  O  N  M  K  J  II    represent    the    octagon 
lion  of  the  roof  and  A'  a  c  C'  the  corresponding  plan.      spire  and   H'  .1'  K'  M'  N'  O'  the  corresponding    plan. 


W 


PLAN 


PLAN. 
Fig.  446.— Plans  and  Elevations  of  Spire.  Fig.  44".-Section  of  Spire  on  Line  of  Hip. 

Pattern  for  an  Octagon  Spire  Mitering  Upon  a  Roof  at  thr  Junction  of  the  Ridge  and  Fli/is. 


Also   let  I)  E  F  G   be    the  side  elevation  of    roof,  and   |   In    the   side  elevation   tlic  spire  is  represented  by  ?  X 
\"  a'  c'  C"  the  corresponding  plan.      In   the  front  ele-      U  T  R  Q,  and    iji    plan    by    Iv  Q'  R'  T"  U'  V.      Only 


Pattern  Problems. 


240 


the  points  in  plans  are  designated  by  letters  which 
represent  similar  points  in  the  elevation.  In  order  to 
draw  the  plans  and  elevations,  including  the  miter 
lines,  it  maybe  found  convenient  to  first  construct  the 
entire  octagons,  as  indicated  in  the  plans,  and  from 
these  to  project  the  elevations  aliove,  as  shown.  From 
the  point  u  in  front  elevation,  which  represents  the 
intersection  of  one  of  the  rear  angles  of  the  spire  with 
the  roof,  carry  a  line  parallel  with  B  F,  cutting  X  //. 
h'roin  the  point  I'  draw  the  miter  line  U  T,  and  from 
the  points  V  1 1  drop  perpendiculars  to  plan,  cutting 
X'  V  and  X'  U',  from  which  points  can  be  drawn 
the  miter  lines  V  IT'  T'  of  the  plan. 

To  obtain  the  miter  line  P'  Q'  It'  of  plan,  from 
which  is  obtained  the  miter  line  Q  R  of  side  elevation, 
a  diagram  has  been  constructed  in  Fig.  447  which 
shows  a  section  of  spire  and  roof  on  the  line  C"  X'  of 
plan.  To  construct  the  diagram  proceed  as  follows : 
Draw  any  line,  as  X°  Y°.  From  X°  set  off  the  dis- 
tance X  E  of  side  elevation  or  W  B  of  front  elevation. 
The  point  E°  represents  the  junction  of  hip  and 
ridge.  From  X°  set  off  the  distance  X  S,  and  erect 
the  perpendicular  S°  L°,  making  it  in  length  equal  to 
S  P,  and  connect  L°  X°.  Then  X°  S°  L°  is  a  duplicate 
of  X  S  P.  From  X°  set  off  the  distance  X  Y  and 
erect  the  perpendicular  Y°  C°,  in  length  equal  to 
X'  C"  of  plan,  and  connect  C°  E°.  Then  L°  X°  repre- 
sents one  side  of  spire,  and  C°  K°  the  hip  of  the  roof, 
and  the  point  Q°  the  point  of  junction  between  the  two. 

As  the  spire  is  a  perfect  octagon,  the  profile  of 
ilie  side  just  constructed  is  in  nowise  different  from 
either  of  those  shown  in  the  elevations.  It  simply  has 
in  addition  the  profile  of  one  of  the  hips  by  means  of 
which  the  correct  hight  of  its  intersection  with  the 


same  (the  point  Q°)  is  determined.  Draw  Q°  Z°  paral- 
lel with  C°  Y°,  and  from  the  point  X'  of  plan  set  off 
the  distance  /J  Q°  of  diagram,  as  shown  by  X'  Q'. 
Connect  R'  Q'  ami  Q'  P'.  From  the  point  Q'  in  plan 
carry  a  line  parallel  with  the  center  line  X  X',  cutting 
the  hip  line  D  E  at  Q.  Draw  Q  R,  which  shows  the 
miter  line  in  side  elevation.  From  the  point  Q  can  be 
drawn  the  line  Q  J,  cutting  the  hip  lines  A  B  and  B  C 
in  front  elevation  at  the  points  J  and  N,  and  the  miter 
lines  J  K  and  M  N  drawn.  The  points  K  M  in  front 
elevation  correspond  with  the  points  K'  M'  of  plan. 

For  the  pattern  proceed  as  follows:  Draw  I  x  of 
Fig.  448,  equal  to  P  X  of  side  elevation,  and  from  I 
erect  the  perpendiculars  /  p  and  I  p',  equal  in  length 
to  L'  P'  <jf  plan  or  L  M  of  front  elevation.  From  y 
and  p'  draw  lines  to  x,  as  shown.  From  x  as  center, 
with  x  p  as  radius,  describe  an  arc,  as  shown  by  p'  e, 
indefinite  in  length.  Set  the  dividers  to  the  distance 
p  p  and  step  off  spaces  from  p,  as  p  r,  r  t,  etc.,  un- 
til as  many  sides  are  set  off  as  are  desired  to  be  shown 
in  one  part  of  the  pattern.  For  convenience  in  de- 
scribing the  pattern  draw  the  lines  x  r,  x  t,  x  c.  Con- 
nect c  and  e  and  m'ake  c  d  equal  to  p  I  and  draw  x  d. 
Bisect  p  r  and  draw  x  a,  and  from  x,  onxa,  set  off  the 
distance  X°  Q°  of  Fig.  447,  locating  the  point  q.  Draw 
p  q  and  q  r.  From  x,  on  x  c?,  set  off  the  distances  X 
V  and  X  U  of  side  elevation,  locating  the  points  v  and 
i.  Through  i  draw  a  perpendicular  cutting  x  c  and 
x  e  in  the  points  u  and  u',  then  draw  u'  v  v  u  and 
u  t.  Then  x  p  p  q  r  t  u  v  u'  is  the  pattern  for  part  of 
spire  shown  on  plan  by  X'  I  P'  Q'  R'  T'  U'  V  I'. 

Fig.  448  shows  a  little  more  than  half  the  full 
pattern,  which  will  be  readily  understood  by  a  com- 
parison of  reference  letters. 


PROBLEM  122. 


The  Envelope  of  a  Rig;ht  Cone. 


In  Fig.  449  let  A  B  C  be  the  elevation  of  the 
cone  and  D  E  F  the  plan  of  the  same.  To  obtain 
the  envelope  set  the  compasses  to  the  space  B  A,  or 
the  slant  hight  of  the  cone,  as  a  radius,  and  from 


any  convenient  point  as  center,  as  B'  of  Fig.  450, 
strike  an  arc  indefinitely.  Connect  one  end  of  the 
arc  with  the  center,  as  shown  by  A'  B'. 

With  the  dividers,  using  as  small  a  space  as  is 


250 


T/ie  Arew  Metal    Worker  Pattern  Book. 


convenient,   step  off  the  circumference    of    the    plan 
D  E  F,  counting  the  spaces  until  the  whole,  or  exactly 


B1  A'  C1  will  be  the  pattern  for  the  envelope  of  the 
cone  ABC. 

It  is  not  necessary  that  all  of  the  spaces  used  in 
measuring  the  circumference  of  the  plan  should  be 
equal.  It  frequently  happens  that  when  the  space 
assumed  between  the  points  of  the  dividers  has  been 
stepped  off  upon  the  circumference  of  the  base,  a  space 
will  remain  at  the  finish  smaller  than  that  original! v 


' 


Fig.  449.— Plan  and  Elevation. 


FiK.  4.VI.  -Pattern. 


The  Envelope  of  a  Right  Cone. 


one  half,  is  completed,  as  shown  in  the  upper  half  of 
the  plan.  Then  set  off  on  the  arc  A1  C'  of  the  pattern, 
commencing  at  A',  the  same  number  of  spaces  as  is 
contained  in  the-  entire  circumference  of  the  plan. 
Connect  the  last  point  C'  with  the  center  B1.  Then 


assumed.  In  that  case  the  required  number  of  full 
spaces  can  be  stepped  off  upon  the  arc  of  the  pattern, 
'after  which  the  remaining  small  space  may  be  added. 
thus  completing  the  correct  measurement  of  the 
pattern. 


PROBLEM  123. 


The  Envelope  of  a  Frustum  of  a  Right  Cone. 


The  principle  involved  in  cutting  the  pattern  for 
the  frustum  of  a  cone  is  precisely  the  same  as  that  for 
cutting  thfe  envelope  of  the  cone  itself.  The  frustum 
of  a  right  cone  is  a  shape  which  enters  so  extensively 
into  articles  of  tinware  that  an  ordinary  flaring  pan,  an 
elevation  and  plan  of  which  are  shown  in  Fig.  451,  has 
been  engraved  for  the  purpose  of  illustration.  An  in- 


spection of  the  engraving  will  show  that  C  D,  the  top 
of  the  pan,  is  the  base  of  an  inverted  cone,  its  apex  ?> 
being  at  the  intersection  of  the  lines  D  0  and  C  A 
forming  the  sides  of  the  pan ;  and  that  A  D  is  the  top 
of  the  frustum  or  the  base  of  another  cone,  A  O  B. 
which  remains  after  cutting  the  frustum  from  the  orig- 
inal cone.  For  the  pattern  then  proceed  as  follows : 


Pattern  Problems. 


251 


Through  the  elevation  draw  a  center  line,  K  B,  indef- 
initely. Extend  one  of  the  sides  of  tho-  pan,  as,  for 
example,  D  O,  until  it  meets  the  center  line  in  the 
point  B.  Still  greater  accuracy  will  be  insured  by  ex- 
tending the  opposite  side  of  the  pan  also,  as  shown — 
the  three  lines  meeting  in  the  point  B — which  deter- 
mines the  apex  of  the  cone  to  a  certainty.  Then  B  0 
and  B  D,  respectively,  are  the  radii  of  the  ares  which 
contain  the  pattern.  From  B  or  any  other  convenient 
point  as  center,  with  B  0  as  radius,  strike  the  arc  P 
Q  indefinitely,  and  likewise  from  the  same  center, 
with  B  D  as  radius,  strike  the  arc  E  F  indef- 
initely. From  the  center  B  draw  a  line  across  these 
arcs  near  one  end,  as  P  E,  which  will  be  an  end 
of  the  pattern.  By  inspection  and  measurement  of  the 
plan  determine  in  how  many  pieces  the  pan  is  to  be 
constructed  and  divide  the  circumference  of  the  pan 
into  a  corresponding  number  of  equal  parts,  in  this  case 
three,  as  shown  by  K,  M  and  L.  With  the  dividers  or 
spacers  step  off  the  length  of  one  of  these  parts,  as 
shown  from  M  to  L,  and  set  off  a  corresponding  num- 
ber of  spaces  on  the  arc  E  F,  as  shown.  Through  the 
last  division  draw  a  line  across  the  arcs  toward  the 
center  B,  as  shown  by  F  Q.  Then  P  Q  F  E  will  be 
the  pattern  of  one  of  the  sections  of  the  pan,  as  shown 
in  the  plan. 


Fi'j.  451. — The  Envelope  of  the  Frustum  of  a  Right  Cone. 


PROBLEM  124. 


To  Construct  a  Ball  in  any  Number  of  Pieces,  of  the  Shape  of  Zones. 


Tn  Fig.  452,  let  A  I  G  II  be  the  elevation  of  a' 
ball  which  it  is  required  to  construct  in  thirteen  pieces. 
Divide  the  profile  into  the  required  sections,  as  shown 
by  0,  1,  2,  3,  4,  etc.,  and  through  the  points  thus  ob- 
tained draw  parallel  horizontal  lines,  as  shown.  The 
divisions  in  the  profile  are  to  be  obtained  by  the  fol- 
lowing general  rule,  applicable  in  all  such  cases : 
Divide  the  whole  circumference  of  the  ball  into  a  num- 
ber of  parts  equal  to  two  times  one  less  than  the  num- 
ber of  pieces  of  which  it  is  to  be  composed. 

In  convenient  proximity  to  the  elevation,  the 
center  being  located  in  the  same  vertical  line  A  N, 
draw  a  plan  of  the  ball,  as  shown  by  K  M  L  N.  Draw 
the  diameter  K  L  parallel  to  the  lines  of  division  in  the 
elevation,  With  the  -square  placed  at  right  angles 


to  this  diameter,  and  brought  successively  against  the 
points  in  the  elevation, drop  corresponding  points  upon 
it,  as  shown  by  1,  2,  3,  4,  etc.  Through  each  of  these 
points  describe  circles  from  the  center  by  which  the 
plan  is  drawn.  Each  of  these  circles  becomes  the 
plan  of  one  edge  of  the  belt  in  the  elevation  to  which 
it  corresponds  in  number,  and  is  to  be  used  in  estab- 
lishing the  length  of  the  arc  forming  the  pattern  of 
the  zone  of  which  it  is  the  base.  Extend  the  center 
line  K-  A  in  the  direction  of  0  indefinitely.  Draw 
chords  to  the  several  arcs  into  which  the  profile  has 
been  divided,  which  produce  until  they  cut  G  A  0,  as 
shown  by  1  2  E,  2  3  D,  3  4  C,  4  5  B  and  5  6  A.  Then 
E  2  and  E  1  are  the  radii  of  parallel  arcs  which  will 
describe  the  pattern  of  the  first  division  above  the  cen- 


252 


The  New  Metal   Worker  Pattern  Book. 


ter  zone,  and  D  3  and  D  2  are  the  radii  describing  the 
pattern  of  the  second  zone,  and  so  on. 

From  El  in  Pig.  453  as  center,  with  E  2  and  E  1 
as  radii,  strike  the  arcs  2  2  and  1  1  indefinitely.    Step 


the  centers  D1,  Fig.  454;  C1,  Fig.  455;  Bl,  Fig.  456, 
and  A',  Fig.  457.  The  pattern  for  the  smallest  section, 
as  indicated  by  F  in  the  plan,  may  be  struck  by  a 
radius  equal  to  F  6  in  the  plan.  The  center  belt  or 


Fig.  455.— Pattern  of  Zone  3 1. 


.  453.— Pattern  of  Zone  1  2. 


Fig.  456.— Pattern  of  Zone  4  i. 


Pig.  457.— Pattern  of  Zone  5  6. 


Fig.  452.— Plan  and  Elevation. 


Fig.  454.— Pattern  of  Zone  2  3.  Fig.  458.— Pattern  of  Middle  Zone. 

To  Construct  a  Ball  in  any  Number  of  Zones. 


off  the  length  on  the  corresponding  plan  line,  and  make 
1  1  equal  to  the  whole  of  it,  or  a  part,  as  may  be  de- 
sired— in  this  case  a  half.  In  like  manner  describe 
patterns  for  the  other  pieces,  as  shown,  struck  from 


zone,  shown  in  the  profile  by  1  0,  is  a  flat  band,  and 
is  therefore  bounded  by  straight  parallel  lines,  the 
width  being  1  0  in  the  elevation,  and  the  length  meas- 
ured upon  line  1  of  the  plan,  all  as  shown  in  Fig.  458. 


Pattern  Problems. 
PROBLEM   125. 
The  Patterns  for  a  Semicircular  Pipe  with  Longitudinal  Seams. 


253 


By  the  nature  of  the  problem  the  pipe  resolves  it- 
self, with  respect  to  its  seetioii  or  profile,  into  some 
regular  polygon.  In  the  illustration  presented  in  Fig. 
•iait  an  oetagoual  form  is  employed,  but  any  other  reg- 


the  profile  ABC  D  F  II  G  E  and  project  the  points 
15  and  C  back  upon  N  R  and  complete  the  elevation  by 
drawing  the  semicircles  O  U  and  P  T. 

By  inspection  of  the  diagram  it  is  evident  that  the 
pattern  for  the  sections  corresponding 
to  0  U  T  P  in  the  elevation  may  be 
pricked  directly  from  the  drawing  as  it 
is  now  constructed,  and  that  the  pat- 
terns for  the  sections  represented  by  E 
A  and  D  F  of  the  profile  will  be  plain 
straight  strips  of  the  width  of  one  side 
of  the  figure,  as  shown  by  either  E  A  or 
D  F,  and  in  length  corresponding  to  the 
length  of  the  sweep  of  the  elevation  on 
the  lines  N  L  V  and  R  X  S,  respectively. 
By  virtue  of  the  bevel  or  flare  of 
the  pieces  N  L  V  U  T  and  R  X  S  T 
P,  as  shown  by  A  B  and  C  D  of 


Fig.  459.— Elevation  and  Section. 


Fig.  480.— Pattern. 


A  Semicircular  Pipe  with  Longitudinal  Seams. 


ular  shape  may  be  used,  and  the  patterns  for  it  will  be 
cut  by  the  same  rule  as  here  explained.  Let  N  L  V  be 
any  semicircle  around  which  an  octagonal  pipe  is  to  be 
curried.  Draw  N  V,  passing  through  the  center  W. 
Through  W  draw  the  perpendicular  L  K  indefinitely. 
Let  N  R  be  the  required  diameter  of  the  octagon. 
Immediately  below  and  iu  line  with  N  R  construct 


the  profile,  each  becomes  one-half  of  the  frustum  of  a 
right  cone,  with  its  apex  above  or  below  the  point  W. 
Therefore  prolong  C  D  of  the  profile  until  it  cuts  the 
center  line  L  K  of  the  elevation  in  the  point  M.  Then 
M  D  and  M  C  are  the  radii  of  the  pieces  corresponding 
to  P  T  S  R  of  the  elevation.  Also  prolong  the  side  A  B, 
or,  for  greater  convenience,  its  equivalent,  E  G,  until  it 


254 


The  X'n-  M'i'tl    Worker  Pattern   Book. 


cuts  the  center  line  in  the  point  M'.  Then  M'  G  and 
M'  E  are  the  radii  of  the  pieces  corresponding  to  N  L  V 
U  O  of  the  elevation.  From  M''  in  Fig.  460  as  center, 
using  each  of  the  several  radii  in  turn,  strike  arcs  in- 
definitely, as  shown  by  N'  V,  O'  U1,  P1  T'  and  K1  S'. 
Step  off  the  length  N  L  V  in  the  elevation,  Fig.  457, 
and  makeN1  V  of  Fig.  45S  equal  to  it.  Draw  N'  O' 
and  V  U'  radial  to  W.  Then  N1  V  U1  O1  will  con- 
stitute the  pattern  for  the  pieces  N  L  V  U  O  of  the 
elevation.  In  like  manner  establish  the  length  of  P' 
T1,  and  draw  P'  R1  and  T1  S1,  also  radial  to  the  center, 
as  shown.  Then  P1  T1  S1  R1  will  be  the  pattern  for 
the  pieces  P  T  S  X  R  of  the  elevation. 

This  rule  may  be  employed  for  carrying  any  polyg- 
onal shape  around  any  curve  which  is  the  segment  of  a 


circle.  The  essential  points  to  be  observed  arc  the 
placing  of  the  profile  in  correct  relationship  to  the  ele- 
vation and  to  the  central  line  L  K,  after  which  prolong 
tin-  oblique  sides  until  they  cut  the  central  line,  thus 
establishing  the  radii  by  which  their  patterns  may  be 
struck.  In  the  case  of  elliptical  curves,  by  resolving 
them  into  segments  of  circles  and  applying  this  rule 
to  each  segment,  as  though  it  were  to  be  constructed 
alone  and  distinct  from  the  others,  no  difficulty  will  be 
met  in  describing  patterns  by  the  principles  here  set 
forth.  The  several  sections  may  be  united  so  as  to 
produce  a  pattern  in  one  piece  by  joining  them  upon 
their  radial  lines.  This  principle  is  further  explained 
in  the  pattern  for  the  curved  molding  in  an  elliptical 
window  cap  in  Problem  128. 


PROBLEM  126. 

The  Blank  for  a  Curved  Molding. 


As  curved  moldings  necessitate  a  stretching  of  the 
metal  in  order  to  accommodate  them  to  both  the  curve 
of  the  elevation  or  plan  and  the  curve  of  the  profile  at 


machinery  designed  for  that  purpose,  care  being 
taken  to  make  the  widtli  of  the  flaring  strip  suffi- 
cient to  include  the  stretchout  of  the  curve  of  the  pro- 


Fiy.  461.— Obtaining  the  Blank  for  a  Curved 
Core  or  Ovolo  Molding. 


Fig.  462.— Obtaining  the  Blank  for  a  Curved 
Ogee  Moldiny. 


the  same  time,  the  patterns  for  their  blanks  can  only 
be  considered  as  flaring  strips  of  metal  in  which  the 
curve  of  the  elevation  or  plan  only  is  considered.  The 
curve  of  the  profile  requires  to  be  forced  into  them  by 


file.  Blanks  for  curved  moldings  thus  become  frus- 
tums of  cones  and  are  cut  according  to  the  principles 
of  regular  flaring  articles,  as  explained  in  the  preceding 
problems.  The  method  of  determining  the  exact  flare 


Pattern   Problems. 


255 


necessary  to  ]>ro<luco  a  certain  mold  with  tin-  greatest 
facility  is  a  matter  to  be  determined  l>y  tho  nature  of 
the  profile  and  tho  kind  of  machinerv  to  lie  used  in 
forming  the  same.  I'sually  a  line  is  drawn  through 
the  extremities  of  the  profile,  as  shown  at  A  D  in 
either  of  the  two  illustrations  here  given,  Figs.  401 
and  -t(i2,  and  is  continued  until  it  meets  the  center  line, 
for  length  of  radius,  as  shown  at  F. 

Therefore,  to  describe  the  pattern  of  the  blank 
from  which  to  make  a  curved  molding  corresponding 
to  the  elevation  AGED,  proceed  in  the  same  manner 


as  though  the  side  E  C  were  to  be  straight.  Through 
the  center  of  the  article  draw  the  line  B  F  indefinitely, 
and  draw  a  line  through  the  points  C  and  E  of  one  of 
the  sides,  which  produce  until  it  meets  B  F  in  the 
point  F.  Then  F  E  will  be  the  radius  of  the  inside  of 
the  pattern.  The  radius  of  the  outside  is  to  be  ob- 
tained by  increasing  F  C  an  amount  equal  to  the  excess 
of  the  curved  line  E  C  over  the  straight  line  E  C,  as 
shown  by  the  distance  C  S.  Then  F  S  is  the  radius  of 
the  outside  of  the  pattern.  The  length  of  the  pattern 
can  be  obtained  as  in  previous  problems. 


PROBLEM  127. 

The  Patterns  for  Simple  Curved  Moldings  in  a  Window  Cap. 


In  Fig.  4ii.'{  is  shown  the  elevation  of  a  window 
cap.  in  the  construction  of  which  two  curved  moldings 
are  required  of  the  same  profile,  but  curved  in  opposite 


essary  for  joining  it  to  the  face  and  roof  pieces  will  be 
obtained  in  one  piece.  The  method  of  developing  the 
pattern  for  the  blank  is  the  same  for  both  curves.  The 


II    10      3 

Fig.  464.— Blank  for  Center  Piece. 
B' 


Fig.  403.— Elevation  or  Window  Cap. 


Fig.  465.  -Blank  for  Side  Piece. 


Thf  Patterns  for  Simple  Curved  Moldings  in  a  Window  Cap. 


directions.  It  is  advisable  to  include  as  much  in  one 
piece  as  can  be  raised  conveniently  with  the  means  at 
hand;  therefore,  the  curved  part  of  the  profile  with  its 
lillets  or  straight  parts  adjacent  and  the  two  edges  nee- 


two  pieces  will  raise  to  the  form  by  the  same  dies  or 
rolls,  it  being  necessary  only  to  reverse  them  in  the 
machine.  Before  the  blank  for  the  middle  piece  can 
be  developed  it  will  be  necessary  to  first  construct  a 


256 


file  New  Metal   Worker  Pattern  Book. 


section  upon  the  center  line,  as  shown  atS  K;  from  all 
points  in  the  mold  and  the  center  of  the  curve  upon  the 
center  line  project  horizontal  lines  to  the  right.  Draw 
any  vertical  line,  as  H  K,  to  represent  the  face  of  the 
cap  in  the  section  and  at  S  draw  the  profile  of  the 
mold,  as  shown.  The  principle  to  be  employed  in  strik- 
ing the  pattern  is  simply  that  which  would  be  used  in 
obtaining  the  envelope  of  the  frustum  of  a  cone  of 
which  A  D  is  the  axis. 

The  general  average  of  the  profile  is  to  be  taken 
in  establishing  the  taper  of  the  cone,  or,  in  other 
words,  a  line  is  passed  through  its  extreme  points. 
Draw  a  line  through  the  profile  in  this  manner  and  pro- 
long it  until  it  intersects  A  D  in  the  point  A,  all  as 
shown  by  C  A.  Then  A  is  the  apex  of  the  cone,  of 
which  A  C  is  the  side  and  H  D  the  top  of  the  frustum. 
Divide  the  profile  S,  as  in  ordinary  practice  for  stretch- 
outs, into  any  number  of  spaces,  all  as  shown  by  the 
small  figures.  Transfer  the  stretchout  of  the  profile  S 
on  to  the  line  A  C,  commencing  at  the  point  1,  as 
shown,  letting  the  extra  width  extend  in  the  direction 
of  C.  From  any  convenient  center,  as  A  in  Fig.  464, 
with  radius  A  C,  describe  the  pattern,  making  the 
length  of  the  arc  equal  to  the  length  of  the  correspond- 
ing'arc  in  the  elevation,  all  as  shown  by  the  spaces  and 
numbers.  From  the  same  center  draw  arcs  correspond- 


ing to  points  9,  10  and  11  of  the  stretchout,  thus  com- 
pleting this  pattern. 

For  the  pattern  of  the  carved  molding  forming  the 
end  portion  of  the  cap  proceed  in  the  same  general 
manner.  Upon  any  line  drawn  through  the  center  N 
of  the  curve,  as  L  M,  construct  a  section  of  the  mold. 
as  shown  at  R.  From  N  draw  the  perpendicular  X  !> 
indefinitely.  Through  the  average  of  the  profile  R,  as 
before  explained,  draw  the  line  to  B,  cutting  N  B  in 
the  point  B,  as  shown.  Lav  off  the  stretchout  of  the 
profile  upon  this  line,  commencing  at  the  point  1,  in 
the  same  manner  as  explained  in  the  previous  operation. 
From  any  convenient  point,  as  B'  in  Fig.  465,  as  center, 
with  radius  B  1,  describe  the  inner  curve  of  the  pat- 
tern, as  shown,  which  in  length  make  equal  to  the  eleva- 
tion, measuring  upon  the  are  1,  all  as  shown  by  the 
small  figures,  after  which  add  the  outer  curves,  as 
shown  by  E1  E". 

The  straight  portion  forming  the  end  of  this  mold- 
ing, as  shown  in  the  elevation,  is  added  by  drawing,  at 
right  angles  to  the  line  E5  B1,  a  continuation  of  the 
lines  of  the  molding  of  the  required  length,  as  shown 
in  the  pattern.  Upon  this  end  of  the  pattern  a 
square  miter  is  to  be  cut  by  the  ordinary  rule  for 
such  purposes,  to  join  to  the  return  at  the  end  of  the 
cap. 


PROBLEM    128. 

The  Pattern  for  the  Curved  Molding:  in  an  Elliptical  Window  Cap. 


In  Fig.  466  is  shown  the  elevation  and  vertical  sec- 
tion of  a  window  cap  elliptical  in  shape,  the  face  of  which 
is  molded.  In  drawing  the  elevation  such  centers  have 
been  employed  as  will  produce  the  nearest  approach  to 
a  true  ellipse  after  the  manner  described  in  Problem  76 
of  Geometrical  Problems,  page  65.  The  centers  B,  D  and 
F,  from  which  the  respective  segments  of  the  elevation 
have  been  described,  may  then  be  used  in  obtaining  pat- 
terns as  follows  :  Through  the  center  F,  from  which 
the  arc  forming  the  middle  part  of  the  cap  is  drawn, 
and  at  right  angles  to  the  center  line  of  the  cap  G  H,  draw 
the  line  I  K  indefinitely.  Project  a  section  on  the  center 
line  of  the  cap,  as  shown  by  P  K  at  the  right,  the  line 
P  K  being  used  as  a  common  basis  of  measurement  upon 
which  to  set  off  the  semi-diameters  of  the  various  cones 
of  which  the  blanks  for  the  moldings  form  a  part. 
Through  the  average  of  the  profile,  as  indicated,  draw 
S  R,  producing  the  line  until  it  meets  I  K.  Divide  the 
profile  of  the  molding  in  the  usual  manner  and  lay  off 


the  stretchout,  as  indicated  by  the  small  figures.  Then 
R  S  is  the  radius  of  the  pattern  of  the  middle  segment 
of  the  cap. 

With  the  dividers,  measuring  down  from  the  pro- 
file, lay  off  on  P  K  distances  equal  to  the  length  of 
the  radius  A  B,  as  shown  by  the  point  O,  and  of  C  1), 
as  shown  by  the  point  M.  Through  these  (mints  () 
and  M,  at  right  angles  to  P  K,  draw  lines  cutting  S  R 
in  the  points  T  and  U.  Then  U  S  is  the  radius  for 
the  pattern  of  the  segment  C  E  of  the  elevation,  and  T  S 
the  radius  of  the  pattern  for  the  segment  A  C.  In 
order  to  obtain  the  correct  length  of  the  pattern,  not 
only  as  regards  the  whole  piece,  but  also  as  regards 
the  length  of  each  arc  constituting  the  curve,  step  off 
the  length  of  the  curved  molding  with  the  dividers 
upon  any  line  of  the  elevation  most  convenient,  as 
shown,  numbering  the  spaces  as  indicated,  and  setting 
off  a  like  number  of  spaces  upon  a  corresponding  line 
of  the  pattern.  As  a  matter  both  of  convenience  ami 


Pattern  Problems. 


257 


accuracy,  the  spaces  used  in  measuring  the  arcs  arc 
greater  in  the  one  of  longest  radius  and  are  diminished 
in  those  of  shorter  radii,  as  will  be  noticed  by  examina- 
tion of  the  diagram. 

To  lay  off  the  pattern  after  the  radii  are  obtained 
as  al>o\-e  described,  proceed  as  follows:  Draw  any 
straight  line,  as  G'  II'  in  Fig.  467,  from  any  point 
in  which,  as  F1,  with  radius  equal  to  R  S,  as  shown 
1'V  F1  E',  describe  an  arc,  as  shown  by  E'G';  and 
likewise,  from  the  same  center,  describe  other  arcs  cor- 
responding to  other  points  in  the  stretchout  of  the  pro- 
iile.  Make  the  length  of  the  arc  E'  G1  equal  to  the 
length  of  the  corresponding  arc  in  the  elevation,  as  de- 
scribed above.  From  E'  to  the  center  F1,  by  which  this 
arc  was  struck,  draw  E'  F'.  Set  the  dividers  to  the 
distance  U  S  as  radius,  with  which,  measuring  from 


length  of  the  corresponding  arc  in  the  elevation, 
all  as  shown  by  the  small  figures.  From  C1  draw  the 
line  C1  D1  to  the  center  by  which  this  arc  was  struck. 


Fig.  467.-Blank  for  the 
Curved  Molding. 


Fig.  486.— Elevation  and  Section  of  Window  Cap.' 

The  Pattern  for  the  Curved  Molding  in  an  Elliptical  Window  Cap. 


along  the  line  E'  F1,  establish  D'  as  center,  from  which 
describe  arcs  corresponding  to  the  points  in  the  profile, 
as  shown  from  E'  to  C'.  Make  E'  C1  equal  to  the 


Set  the  dividers  to  the  distance  T  S  in  the  section, 
and,  measuring  from  C'  along  the  line  C1  D1,  establish 
the  point  B1,  from  which  as  a  center  strike  arcs  cor- 


258 


The  Xew  Metal    Worker  Pattern  B<j<jk. 


responding  to  those  already  described  in  the  other 
section  of  the  pattern.  Make  the  length  equal  to  the 
length  of  the  corresponding  segments  in  the  elevation, 


and  draw  the  line  A'  IV.  Then  A'  C1  E'  G'  is 
the  half  pattern  corresponding  to  A  0  E  G  of  the 
elevation. 


PROBLEM  129. 


The  Pattern  of  an  Oblong  Raised  Cover  with  Semicircular  Ends. 


In  Fig.  468  let  A  B  C  D  represent  a  side  eleva- 
tion of  the  cover  of  which  E  G  F  H  is  the  plan  or 
shape  of  the  vessel  it  is  to  fit.  Various  constructions 
may  be  employed  in  making  such  a  cover  as  this ; 
that  is,  the  joints,  at  the  option  of  the  mechanic,  may 
be  placed  at  other  points  than  shown  here;  the  prin- 
ciple used  in  obtaining  the  shape,  however,  is  the 
same,  whatever  may  be  the  location  of  the  joints. 
By  inspection  of  the  elevation  and  plan  it  will  be  seen 
that  the  shape  consists  of  the  two  halves  of  the  en- 
velope of  a  right  cone,  joined  by  a  straight  piece. 
Therefore,  for  the  pattern  proceed  as  follows :  At  any 
convenient  point  lay  off  B2  C",  in  length  equal  to  B1 
C'  of  the  plan.  From  B2  and  C"  as  centers,  with 
radius  equal  to  A  B  or  C  D  of  the  elevation,  deseriKe 
arcs,  as  shown  by  0  N  and  P  M.  Upon  these  arcs, 
measured  from  0  and  P,  respectively,  set  off  the 
stretchout  of  the  semicircular  ends,  as  shown  in  plan, 
thus  obtaining  the  points  M  and  N.  From  N  draw 
N  B',  and  from  M  draw  M  C\  From  B3  and  C2,  at, 
right  angles  to  the  line  B"  C'',  draw  B"  K  and  C3  L,  in 
length  equal  to  A  B  of  the  elevation,  which  represents 


Fig.  46S. — Elevation,  Plan  and  1'attern    of  an  Oblong  Raised  Cover 
with  Semicircular  Ends. 


the  slant  hight  of  the  article.  Connect  K  and  L,  as 
shown.  Then  0  N  K  L  M  P  will  be  the  required 
pattern. 


PROBLEM   130. 


The  Pattern  of  a  Regular  Flaring  Article  which  Is  Oblong  with  Semicircular  Ends. 


In  Fig.  469,  let  A  B  D  C  be  the  side  elevation  of 
the  required  article.  Below  it  and  in  line  with  it  draw 
a  plan,  as  shown  by  E  c  d  F  H  G.  From  D  in  the 
elevation  erect  the  perpendicular  D  L.  Then  L  C 
represents  the  flare  of  the  article  and  C  D  is  the  width 
of  the  pattern  throughout.  Across  the  plan,  at  the 
point  where  the  curved  end  joins  the  straight  sides. 
draw  the  line  d  H  at  right  angles  to  the  sides  of  the 
article.  As  the  plan  may  be  drawn  at  any  distance 


from  the  elevation,  this  line  must  be  prolonged,  if 
necessary,  to  meet  C  D  extended.  Produce  C  I)  until 
it  meets  d  H,  as  shown  by  y.  Then  y  D  and  g  C  are 
radii  of  the  curved  parts  of  the  pattern. 

Lay  off  on  a  straight  line  M  O  in  Fig.  470,  the 
length  of  the  straight  part  of  the  article,  as  shown  in 
the  plan  by  c  d.  At  right  angles  to  M  0  draw  M  S 
and  0  K  indefinitely.  Upon  these  lines  set  off  from 
M  and  0  the  distance  g  C,  locating  the  points  S  and  $, 


Pattern  Problems. 


259 


the  centers  for  the  curved  portions  of  the  pattern.. 
From  S  with  the  radius  y  ('  strike  the  are  M  I"  indef- 
initely. In  like  manner,  with  same  radius,  from  R  as 
center  describe  the  arc  0  V.  From  the  same  centers, 
with  radius  equal  to  g  D,  describe  the  arcs  N  T  and  P 
.W.  Step  off  the  length  of  the  curved  part  of  the  article 
upon  either  the  inner  or  outer  line  of  the  plan,  and  make 
the  corresponding  arc  of  the  pattern  equal  to  it,  as 
shown  by  the  spaces  in  N  T  and  1'  W.  Through  the 
points  T  and  W  draw  lines  from  the  centers  S  and  R, 
producing  them  until  they  cut  the  outer  arcs  at  U  and 
V.  At  right  angles  to  the  line  S  T  U  or  K  W  V,  as 
the  case  may  be,  set  off  V  X  Y  W,  equal  to  M  O  P  N", 
which  will  be  the  other  straight  side  of  the  pattern. 
Then  U  M  O  V  X  Y  W  P  N  T  will  be  the  complete 
pattern  in  one  piece. 

If  it  were  desired  to  locate  the  seam  midway  in 


(Mid,  instead  of  all  at  one  end,  as  shown.     In  like  man- 
ner such  changes   may  be  made  as  are  necessary  for 


T 


Fig.  4«9.— Plan  and  Elevation.  Fig.  470.— Pattern. 

The  Pattern  of  a  Regular  Flaring  Article  Which  is  Oblong  with  Semicircular  Ends. 


one  of  the  straight  sections,  in  adding  the  last  member 
as   above  described,  one-half  would  be  placed  at  each 


locating  the  seam  at  any  other  point,  or  for  cutting  the 
pattern  in  as  many  pieces  as  desired. 


PROBLEM    131. 


The  Pattern  of  a  Regular  Flaring  Oblong  Article  with  Round  Corners. 


In  Fig.  471,  A  C  D  B  is  the  side  elevation  of  the 
article  and  E  F  G  M  N  0  P  B  the  plan.  The  corners 
are  arcs  of  circles,  being  struck  by  centers  H,  L,  T 
and  S,  as  shown.  Draw  the  plan  in  line  with  the  ele- 
vation, so  that  the  same  parts  in  the  different  views 


shall  correspond.  Through  the  centers  H  and  L  of 
the  plan  by  which  the  corners  FG  and  M  N  are  struck, 
draw  F  N  indefinitely.  Prolong  the  side  line  of  the 
elevation  C  D  until  it  cuts  F  N  in  the  point  K,  as 
shown.  Then  K  D  is  the  radius  of  the  inside  line  of 


2GO 


TJie  New  Metal   Worker  Pattern  Book. 


the  pattern  of  the  curved  part,  and  K  C  is  the  radius 
of  the  outside  line. 

Draw  the  straight  line  E'  F1  of  Fig.  472,  in  length 
equal  to  the  straight  part  of  one  side  of  the  article,  or  E 
F  of  the  plan.  Through  the  points  E1  and  F',  at  right 
angles  to  the  line  E1  F',  draw  lines  indefinitely,  as  shown 
by  E1  U  and  F1  K1.  Upon  these  lines  set  off,  from  F1  and 
E1,  the  distance  K  C,  locating  the  points  K1  and  U,  the 
centers  for  the  curved  parts.  From  K',  with  the  radius 
K  C,  strike  the  arc  F1  G1,  which  in  length  make  equal 
to  F  G  of  the  plan.  From  G1  draw  a  line  to  the  center 
K1,  at  right  angles  to  which  erect  G'  M1,  in  length 
equal  to  G  M  of  the  plan.  In  like  manner,  with  like 
radius,  describe  the  arc  E1  R'.  Draw  R1  U,  at  right 


right  angles  to  it  lay  off  O1  N1,  equal  to  0  X  of  the 
elevation.     Draw  N'  W,  and  draw  the  arc  N'  M"  in  the 


N 

fig-  471.— Plan  and  Elevation.  Fig.  472.— Pattern. 

The  Pattern  of  a  Regular  Flaring  Oblong  Article  With  Round  Corners. 


angles  to  which  erect  R1  P1,  equal  to  R  P  of  the  plan. 
At  right  angles  to  R1  P1  draw  P1  V  indefinitely.  In 
the  manner  above  described  establish  the  center  V,  and 
from  it  describe  the  third  arc  P'  0'.  Draw  O1  V.  At 


same  manner  as  already  described.  In  the  same  man- 
ner lay  off  the  inner  line  of  the  pattern,  as  shown  by 
in  yfe  r  p  o  n  ra'.  Join  the  ends  M'  m  and  Ma  m1,  1 1m* 
completing  the  pattern  sought. 


PROBLEM  132. 

The  Envelope  of  the  Frustum  of  a  Cone,  the  Base  of  Which  Is  an  Elliptical  Figure. 


This  shape  is  very  frequently  used  in  pans  and 
plates,  and  therefore  in  Fig.  473  is  shown  an  elevation 
and  plan  of  what  is  familiarly  termed  an  oval  flaring 


pan.  Let  that  part  of  the  plan  lying  between  H  and 
L  be  an  arc  whose  center  is  at  U,  and  let  those  por- 
tions between  V  and  II  and  L  and  \\  be  arcs  whose 


Pattern  Problems. 


261 


centers  are,  respectively,  R  and  S.  A  C  D  B  represents 
an  elevation  of  the  vessel,  and  is  so  connected  with  the 
plan  as  to  show  the  relationship  of  corresponding 
points. 

The  first  step  is  to  construct  a  diagram,  shown  in 
Fig.  474,  1>\-  means  of  which  the  lengths  of  the  nidii  to 
be  used  in  describing  the  pattern  are  to  l>e  obtained. 
Draw  the  horizontal  line  II  U  indefinitely,  and  at  right 
angles  to  it  draw  II  A,  indefinitely  also.  Make  II  U 
equal  to  II  U  of  the  plan,  Fig.  471.  Make  II  C  equal 
to  the  vertical  hight  of  the  vessel,  as  shown  in  the 


w 


Fig.  473.— Plan  and  Elevation. 


N  0.  From  the  points  H  and  L  of  the  arc  first  drawn 
draw  lines  to  A,  thus  intercepting  the  arc  N  O  and 
determining  its  length. 

In  the  diagram,  Fig.  474,  set  off  from  II,  on  the 
line  II  U,  the  distance  R  H,  making  it  equal  to  R  II 
of  the  plan,  Fig.  473.  Also,  upon  the  line  C  G,  from 
the  point  C,  set  off  C  I,  equal  to  R  N  of  the  plan. 
Then,  through  the  points  R  and  I  thus  established, 
draw  the  line  R  B,  which  produce  until  it  intersects 
A  H.  Then  R  B  will  be  the  radius  for  those  portions 
of  the  pattern  lying  between  V  and  H  and  L  and  W  of 


Fig.  474.— Diagram  of  Radii. 


Fig.  475.— Pattern  of  One  Half. 


The  Envelope  of  the  Frustum  of  a  Cone,  the  Base  of  Which  is  an  Elliptical  Figure. 


elevation  by  D  X.  Draw  the  line  C  G  parallel  to  II  U, 
making  C  G  in  length  equal  to  U  N  of  the  plan. 
Through  the  points  U  and  G  thus  established  draw  the 
line  U  G,  which  continue  until  it  meets  II  A  in  the 
point  A.  Then  A  U  will  be  the  radius  by  which  to 
describe  that  portion  of  the  pattern  which  is  included 
between  the  points  H  and  L  of  the  plan.  With  A  U 
as  radius,  and  from  any  convenient  point  as  center — as 
A,  Fig.  475 — draw  the  arc  II  L,  which  in  length  make 
equal  to  II  L  of  the  plan,  Fig.  473,  as  shown  by  the 
points  1,  12,  3,  etc.  From  the  same  center,  and  with 
the  radius  A  G  of  Fig.  474,  describe  the  parallel  arc  i 


the  plan.  From  the  point  H,  on  the  line  H  A,  Fig. 
475,  set  off  the  distance  H  B,  equal  to  R  B  of  Fig. 
474.  Then,  with  B  as  center,  describe  the  arc  E  II, 
and  from  a  corresponding  center,  C,  at  the  opposite  end 
on  pattern,  describe  the  arc  L  K.  'From  the  same 
centers,  with  B  I  as  radius,  describe  the  arcs  N  M  and 
O  P,  all  as  shown.  Make  H  E  and  L  K  in  length 
equal  to  II  E  and  L  K  of  the  plan.  From  E  and  K, 
respectively,  draw  lines  to  the  centers  B  and  C,  inter- 
cepting the  arcs  N  M  and  0  P  in  the  points  M  and  P. 
Then  E  K  P  M  will  be  one-half  of  the  complete  pat- 
terns of  the  vessel. 


PROBLEM  133. 
The  Pattern  of  a  Heart-Shaped  Flaring  Tray. 


Let  E  C  G1  F  G  C1  of  Fig.  476  be  the  plan  of  the 
article,  and  I  N  O  K  the  elevation.  By  inspection  of 
the  plan  it  will  be  seen  that  each  half  of  it  consists  of 


two  arcs,  one  being  struck  from  D  or  D1  as  center,  and 
the  other  from  C  or  C1  as  center,  the  junction  between 
the  two  arcs  being  at  G  and  G',  respectively.  From  C 


262 


The  Neiu  Metal   Worker  Pattern  Book. 


draw  C  F,  and   likewise  draw  C  G.     Upon    the   point 
D1  erect  the  perpendicular  D1  C'. 

To  obtain  the  radii  of  the  pattern  construct  a  dia- 


Lay  off  the  perpendiculars  X  U  and  P  $  indefinitely. 
Upon  P  8,  from  P,  set  off  P  R,  equal  to  1)'  C1  of  the 
plan,  and  on  X  U,  from  X,  set  off  X  W,  equal  to  D'  c  of 


Fix.  476.— Plan  and  Elevation. 


Fig.  478.— Pattern. 

The  Pattern  of  a  Heart-Shaped  Flaring  Tray. 


n 
R.  477.— Diagram  of  Uadii. 


gram,  shown  in  Fig.  477,  which  is  in  reality  a  section 
upon  the  line  C  G  of  the  plan.  Draw  X  P  in  Fig.  477, 
in  length  equal  to  the  straight  hight  of  the  article. 


the  plan.  In  like  manner  make  P  S  equal  to  C  G  of 
the  plan,  and  X  U  equal  to  C  .7  of  the  plan.  Connect 
U  S  and  W  R.  Produce  P  X  indefinitely  in  the  direc- 


Pattern  Problems. 


263 


tion  of  Z.  Also  produce  R  W  until  it  meets  P  X  in 
tin-  point  Y.  and  in  like  manner  produce  S  I"  until  'it 
meets  P  Z  in  the  point  Z.  Tlien  /  I"  and  'A  S  are  the 
radii  for  that  portion  of  the  article  contained  between 
G  and  K  of  the  plan,  and  V  \V  and  Y  Rare  the  radii  of 
that  portion  shown  from  (i  to  E  of  the  plan. 

To  lay  out  the  pattern  after  the  radii  are  estab- 
lished, draw  anv  straight  line,  as  Z'  G*  in  Fig.  47s,  in 
length  equal  to  Z  S  of  the  diagram.  From  Z'  as  cen- 
ter, with  Z  S  as  radius,  describe  the  arc  G*  F1,  in  length 


ei[tial  to  G  F  of  the  plan.  In  like  manner,  with  radius 
Z  I",  from  the  same  center,  describe  the  arc  y'f,  in 
length  equal  to #/ of  the  elevation.  Draw/' F'.  Set 
of]'  from  G2,  upon  the  line  G2  Z',  the  distance  R  Y  of 
Fig.  477,  as  shown  at  Y',  and  from  Y1  as  center,  with 
the  radius  R  Y,  describe  the  arc  G2E',  which  in  length 
make  equal  to  G  E  of  the  plan.  In  like  manner,  from 
tin;  same  center,  with  radius  Y  "W,  describe  the  arc 
.</'  '''.  equal  to  the  arc  g  c  of  the  plan.  Draw  c'  E1,  thus 
completing  the  required  pattern. 


PROBLEM    134. 

The  Pattern  of  an  Oval  or  Eg-g-Shaped  Flaring;  Pan. 


Let  A  B  C  D  in  Fig.  479  represent  the  elevation  |  construct  a  section  of  the  article  as  it  would  appear  if 


of  the  article,  of  which  A1  K  L  B'  M  I  is  the  plan. 
The  plan  is  constructed  by  means  of  the  centers  O,  P, 
F  and  F1,  as  indicated.  The  patterns,  therefore,  are 


cut  on  the  line  A1  P  of  the  plan.  Therefore  set  off, 
at  right  angles  to  it,  A2  P3,  equal  to  A1  P.  Make 
P5  D"  equal  to  the  straight  hight  of  the  article,  as 


f'- 


Fig.  4HO.— Diagram  of  Small  Cone. 


Fig.  479. -Plan  and  Elevation. 


Fig.  481.— Diagram  of  Middle  Cone. 
An  Oval  or  Egg-Shaped  Flaring  Pan. 


Fig.  482.— Diagram  of  Large  Cone. 


struck  by  radii  obtained  from  sections  of  the  several 
cones  of  which  the  article  is  composed.  At  any  con- 
venient place  draw  the  line  P2  P',  Fig.  480,  indefi- 
nitely, corresponding  to  P  of  the  plan,  and  upon  it 


shown  by  R  D  of  the  elevation.  Make  D1  A"  of  the 
diagram  equal  to  D1  P  of  the  plan.  Draw  A1  A', 
which  will  correspond  to  A  D  of  the  elevation,  pro- 
longing it  until  it  meets  P2  P'  in  the  point  P1.  Then 


New  Metal   Worker  Pattern 


P1  A*  is  the  radius  of  the  outside  line  of  the  pattern 
of  the  portion  between  K  and  I  of  the  plan,  and  P'  A3 
is  the  radius  of  the  line  inside  of  the  same  part. 

In  like  manner  draw  the  line  O1  O2,  Fig.  481, 
corresponding  to  O  of  the  plan,  and  construct  a  section 
taken  on  the  line  0  B1,  as  shown  by  O2  B2  C2  C3. 
Produce  B2  C2  until  it  meets  O2  O1  in  the  point  O1. 
Then  O1  C2  and  O1  B3  are  the  radii  of  the  pattern  of 
that  portion  of  the  article  contained  between  L  and  M 
of  the  plan. 

Draw  the  line  F3  F2,  Fig.  482,  which  shall  corre- 
spond to  F  or  F1  of  the  plan.  Make  F"  E  equal  to  the 
straight  hight  of  the  article,  and  lay  off  F'  L3  at  right 
angles  to  it,  equal  to  F1  L  of  the  plan,  and  E  Z2  equal 
to  F1 .1  of  the  plan.  Draw  L3  ?,  which  produce  until 
it  meets  F3  F2  in  the  point  F1.  Then  F2  I  and  F2  L!, 
respectively,  are  the  radii  of  the  pattern  of  those  parts 
shown  by  K  L  and  I  M  of  the  plan. 

To  lay  off  the  pattern  after  the  several  radii  are 
obtained,  as  described  above,  draw  any  straight  line, 
in  length  equal  to  F2  L3,  as  shown  by  F4  K  in  Fig. 
483,  and  from  F*  as  center,  with  F2  V  and  F2  L3,  Fig. 
482,  as  radii,  strike  arcs,  as  shown  by  &'  Z1  and  K  L1, 
which  in  length  make  equal  to  the  corresponding  arcs 
of  the  plan  K  L  and  k  I,  as  shown.  Draw  L1  F4,  and 
upon  it  set  off  from  Z1  toward  F4,  a  distance  equal  to 
O1  C'  of  Fig.  481,  establishing  the  point  O3,  from 
which  strike  the  arc  T  m1,  in  length  equal  to  I  m  of 
the  plan.  In  like  manner,  from  the  same  center, 
with  radius  O1  B',  of  Fig.  481,  strike  the  arc  L1  M1, 
equal  in  length  to  L  M  of  the  plan.  Draw  M'  O3, 
which  produce  in  the  direction  of  F6  making  M'  Fb 
equal  to  K  F4,  from  which  center  continue  the  inner 
line  of  the  pattern,  as  shown  by  m  i\  which  in  length 
must  equal  m  i  of  the  plan.  In  like  manner,  from  the 
same  center,  with  radius  F2  L3  of  Fig  482,  describe 
the  arc  M1 11  and  draw  I1  F3.  Set  off  on  this  line  from 
P  a  distance  equal  to  P'  A3  of  Fig.  480,  thus  estab- 
lishing the  center  P3.  Describe  the  arc  t"  k,  in  length 
equal  to  i  k  of  the  plan.  In  like  manner,  from  the 
same  center,  with  the  radius  P1  A2  of  Fig.  480  describe 
the  arc  I1  K",  in  the  length  equal  to  I  K  of  the  plan. 
Place  the  straight  edge  against  the  points  P3  and  K2 
and  draw  K2  k,  thus  completing  the  pattern. 


From  inspection  it  is  evident  that  the  pattern 
might  have  been  commenced  at  any  other  point  as 
well  as  at  K  k  of  the  plan.  If  the  joint  is  desired 
upon  any  of  the  other  divisions  between  the  arcs,  as 
L  I,  M  m,  or  I  i,  the  method  of  obtaining  it  will  be  so 
nearly  the  same  as  above  narrated  as  not  to  require 
special  description.  If  the  joint  is  wanted  at  some 
point  in  one  of  the  arcs  of  the  plan,  as,  for  example, 


Fig.  4S3.—The  Pattern  of  an  Oval  or  E;/g-Shaped  Flarimj  Pan. 

at  X  x,  draw  the  line  X  x  across  the  plan,  producing 
it  until  it  meets  the  center  by  which  that  arc  of  the 
plan  is  struck.  In  laying  off  the  pattern,  commence 
with  a  line  corresponding  to  X  F1,  in  place  of  F4  K', 
and  from  it  lay  off  an  arc  corresponding  to  the  portion 
of  the  arc  in  the  plan  intercepted  by  X  ,r,  as  shown  by 
XL?  x.  Proceed  in  other  respects  tin-  same  as  above 
described  until  the  line  k  K2  is  obtained,  against  which 
there  must  be  added  an  arc  corresponding  to  the 
amount  cut  from  the  first  part  of  the  plan  by  X  a;,  as 
above  described,  or,  in  other  words,  equal  to  X  K  k  x 
of  the  plan. 


Pattern  Problems. 
PROBLEM  135. 

The  Envelope  of  the  Frustum  of  a  Right  Cone,  the  Upper  Plane  of  Which  Is  Oblique  to  Its  Axis. 


265 


In  Fig.  4-S4,  let  C  B  D  E  lie  tin-  elevation  of  the 
required  shape.  Produce  the  sides  C  B  and  K  1)  un- 
til thev  intersect  at  A.  Then  A  will  lie  the  :i]>ex  of 


Fit/.  4*4. — The  Enrelopf  of  th< 


Frustum  of  a  Rif/hf    Cone  Whose  Upper  Plane  is 
Oblique  to  its  Ajrin. 


the  cone  of  which   C  B  D  E  is  a  frustum.      Draw  the 
axis  A  G,  which  produce  below  the  lignre,  and  from  a 


center  lying   in   it   draw  a  half  plan  of  the  article,  as 
shown  by  F  G  II. 

Divide  this  plan  into  any  number  of  equal  parts, 
and  from  the  points  carry  lines 
parallel  to  the  axis  until  they  cut 
the  base  line,  and  from  there  extend 
them  ^in  the  direction  of  the  apex 
until  they  cut  the  upper  plane  B  D. 
Place  the  T-square  at  right  angles 
to  the  axis,  and,  bringing  it  against 
the  several  points  in  the  line  B  D, 
cut  the  side  A  E,  as  shown.  From 
A  as  center,  with  A  E  as  radius, 
describe  the  arc  C'  E',  on  which 
lay  off  a  stretchout  of  either  a  half 
or  the  whole  of  the  plan,  as  may 
be  desired,  in  this  case  a  half,  as 
shown.  From  the  extremities  of 
this  stretchout,  C1  and  E',  draw  lines 
to  the  center,  as  C'  A  and  E1  A. 
Through  the  several  points  in  the 
stretchout  draw  similar  lines  to  the 
center  A,  as  shown.  With  the 
point  of  the  compasses  set  at  A, 
bring  the  pencil  to  the  point  D  in 
the  side  A  E,  and  with  that  radius 
describe  an  arc,  which  produce  until 
it  cuts  the  corresponding  line  in  the 
stretchout,  as  shown  at  D1.  In  like 
manner,  bringing  the  pencil  against 
the  several  points  between  D  and  E 
in  the  elevation,  describe  arcs  cut- 
ting the  corresponding  measuring 
lines  of  the  stretchout.  Then  aline 
(raced  through  these  intersections 
will  form  the  upper  line  of  the 
pattern,  the  pattern  of  the  entire  half  being  con- 
tained in  C1  B'  D'  E1. 


PROBLEM    136. 

The  Envelope  of  a  Right  Cone  Whose  Base  Is  Oblique  to  Its  Axis. 


In  Fig.  485,  let  G  D  H  be  the  elevation  of  u  right 
cone  whose  base  is  oblique  to  its  axis,  the  pattern  of 


which  is  required.     It  will  be  necessary  first  to  assume 
any  section  of  the  cone  at  right  angles  to  its  axis  as  a 


Tlie  Xew  Metal   Worker  Pattern  Book. 


base  upon  which  to  measure  its  circumference.  This 
can  be  taken  at  any  point  above  or  below  the  ol>li<|iu> 
base  according  to  convenience. 

Therefore  at  right  angles  to  the  axis  D  O,  and 
through  the  point  G,  draw  the  line  G  F.  Extend  the 
axis,  n.s  shown  by  D  B,  and  upon  it  draw  a  plan  of 
the  cone  as  it  would  appear  when  cut  upon  the  line  G  F, 
as  shown  bv  ABC.  Divide  the  plan  into  any  convenient 
number  of  equal  parts,  and  from  the  points  thus  ob- 
tained drop  lines  on  to  G  F.  From  the  apex  D, 
through  the  points  in  G  F,  draw  lines  to  the  base 
G  II.  From  D  as  center,  with  D  G  as  radius,  describe 
an  arc  indefinitely,  on  which  lay  off  a  stretchout  taken 
from  the  plan  ABC,  all  as  shown  by  I  M  K.  From 
the  center  D,  by  which  the  arc  was  struck,  through 
the  points  in  the  stretchout,  draw  radial  lines  indefi- 
nitely, as  shown.  Place  the  blade  of  the  T-square 
parallel  to  the  line  G  F,  and,  bringing  it  against  the 
several  points  in  the  base  line,  cut  the  side  D  H,  as 
shown,  from  F  to  II.  With  one  point  of  the  com- 
passes in  D,  bring  the  other  successively  to  the  points 
1,  2,  3,  4,  etc.,  in  F  II,  and  describe  arcs,  which  pro- 
duce until  they  cut  the  corresponding  lines  drawn 
through  the  stretchout,  as  indicated  by  the  dotted 
lines.  Then  a  line,  ILK,  traced  through  these 
points  of  intersection,  as  shown,  will  complete  the  re- 
quired pattern. 


Pattern 


Elevation.    ^£^ 

0 


Fig.  Jj8S.—Thf  Envelope  of  a  Cone  whose  Base  is  Obliqur  to  its  ATI'S 


PROBLEM  137. 
A  Conical  Flange  to  Fit  Around  a  Pipe  and  Against  a  Roof  of  One  Inclination. 


In  Fig.  486  is  shown,  by  means  of  elevation  and 
plan,  the  general  requirements  of  the  problem.  A  B 
represents  the  pitch  of  the  roof,  G  H  K  I  represents  the 
pipe  passing  through  it,  and  C  D  F  E  the  required  flange 
fitting  around  the  pipe  at  the  line  C  D  and  against 
the  roof  at  the  line  E  F.  The  flange,  as  thus  drawn, 
becomes  a  portion  of  the  envelope  of  aright  cone. 

At  any  convenient  distance  below  the  elevation 
assume  a  horizontal  line  as  a  base  of  the  cone  upon 
which  to  measure  its  diameter,  and  continue  the  sides 
downward  till  they  intersect  this  base  line,  all  as  shown 
at  L  M.  Also  continue  the  sides  upward  till  they  in- 
tersect at  W,  the  apex.  Below  the  elevation  is  shown 
a  plan,  and  similar  points  in  both  news  are  connected 
by  the  lines  of  projection.  S  T  represents  the  pipe 
and  N  O  the  flange.  While  the  pipe  is  made  to  pass 
through  the  center  of  the  cone,  as  may  be  seen  by  ex- 
amining the  base  line  L  M  in  the  elevation,  and  also 


P  R  of  the  plan,  it  does  not  pass  through  the  center  of 
the  oblique  cut  E  F  in  the  elevation,  or,  what  is  the 
same,  N  0  of  the  plan. 

For  the  pattern  of  the  flange  proceed  as  shown  in 
Fig.  481,  which  in  the  lettering  of  its  parts  is  made  to 
correspond  with  Fig.  486,  just  described.  Divide  the 
half  plan  P  X  11  into  any  convenient  number  of  parts — in 
this  case  twelve — and  from  each  of  the  points  thus 
established  erect  perpendiculars  to  the  base  of  the  cone, 
obtaining  the  points  1',  :>',  31,  etc.  Fmni  these  points 
draw  lines  to  the  apex  of  the  cone  \V .  cutting  the 
oblique  line  E  F  and  the  top  of  the  flange  ('  1>.  as 
shown.  Inasmuch  as  C  D  cuts  the  cone  at  right  angles 
to  its  axis,  the  line  in  the  pattern  corresponding  to  it 
will  be  an  arc  of  a  circle;  but  with  E  F,  which  cuts  the 
cone  obliquclv  to  its  axis,  the  ease  is  different,  eaeli 
point  in  it  being  at  a  different  distance  from  the  apex. 
Accordingly,  the  several  points  in  E  F,  obtained  by 


Pattern  Problems. 


267 


the  lines  from  the  plan  drawn  to  the  apex  \V,  must  In- 
transferred  to  one  of  the  sides  of  the  cone,  where  their 
distances  from  \Y  can  he  accurately  measured.  There- 
fore from  the  points  0",  I3,  23,  3:i,  in  K  l'\  draw  lines 
ut  right  angles  to  the  axis  of  the  cone  AY  X,  cutting 
the  side  W  M,  as  shown.  With  W  as  center,  and  with 
\V  M  as  radius,  strike  the  arc  1"  11'  indefinitely,  and. 


in  the  plan  P  X  11,  all  as  shown  hy  0',  1",  2',  3",  etc. 
From  these  points  draw  lines  to  the  center  \V,us  shown. 
With  one  point  of  the  dividers  set  at  AY  and  the  other 
brought  successively  to  the  points  obtained  in  AY  M  by 
the  horizontal  lines  drawn  from  E  F,  cut  the  correspond- 
ing lines  in  the  stretchout  of  the  pattern,  as  indicated 
hy  the  curved  dotted  lines.  A  line  traced  through 


w^i 


Fig.  4»6.-Plan  and  Elevation. 


Fig.  487.— Pattern. 
A  Conical  Flange  to  Fit  Around  a  Pipe  and  Against  a  Roof  of  One  Inclination. 

these  points,  as  E1  F',  will  represent  the  lower  side  of 
the  pattern.  As  but  one-half  of  the  plan  has  been 
used  in  laying  out  the  stretchout,  the  pattern  C'  E1  F1 
D'  thus  obtained  is  but  one-half  of  the  piece  required. 
It  can  be  doubled  so  that  the  scam  can  be  made  to 
come  through  the  short  side  at  C  E,  or  through  the 
long  side  at  D  F,  at  pleasure. 


with  the  same  center  and  with  W  D  as  radius,  strike 
the  arc  C1  D'  indefinitely,  which  will  form  the  bound- 
ary of  the  pattern  at  the  top.  At  anj-  convenient  dis- 
tance from  M  draw  AY  P',  a  portion  of  the  length  of 
which  will  form  the  boundary  of  one  end  of  the  pat- 
tern. On  P'  K',  commencing  with  P',  set  off  spaces 
equal  in  length  and  the  same  in  number  as  the  divisions 


PROBLEM  138. 

The  Pattern  for  a  Cracker  Boat. 


LetE  F  H  G,  in  Fig.  4SS,  be  the  side  elevation, 
A  B  C  D  E  the  end,  and  I  K  J  L  the  plan  of  a  dish 
sometimes  called  a  cracker  boat  or  bread  trav.  The 


sides  of  the  dish  are  parts  of  the  frustum  of  a  right 
cone.  To  the  plan  have  been  addc]  the  circles  show- 
ing the  complete  frustums  of  which  the  sides  are  a  part, 


268 


Tlie  New  Metal   Worker  Pattern  Book. 


L  and  K  being  the  centers,  all  of  which  will  appear 
clear  from  an  inspection  of  the  drawing,  and  below  is 
further  shown  a  side  view  of  this  frustum.  While  in 


Fig.  488.-Plan  and  Elevations.  Fig.  489.— Pattern. 

The  Pattern  for  a  Cracker  Boat. 

the  plan  the  top  and  bottom  of  the  sides  have  been 
shown  parallel,  in  the  side  view  the  top  appears  curved 
at  C,  the  cut  producing  which  curve  being  shown  by 
B  C  of  the  end  view. 

Extend  the  sides   IT  "W  and  V  X  of  the  frustum 
until  they  meet  at  Z,  which  is  the  apex  of  the  com- 


pleted cone.  Before  the  pattern  can  be  described  it 
will  be  necessary  to  draw  a  half  elevation  of  the  cone 
U  V  Z,  showing  the  end  view  of  the  tray  in  its  relation 
to  the  same,  as  in  Fig.  480.  Draw  any  center  line,  as 
K'  Z'.  From  the  point  L',  as  center,  strike  the  arc  K' 
T',  being  one-fourth  of  the  plan  of  top,  as  shown  by  K 
Tin  Fig.  488.  Below  the  plan  of  top  draw  one-half 
of  frustum  of  cone,  as  shown  by  «'  V'X'  w' ',  in  which 
draw  the  end  elevation  of  boat  A'  B'  C'  X'  K',  letting 
V  X'  be  one  of  its  sides,  and  extend  the  line  b'  B' 
through  the  arc  K'  T'  at  B".  Divide  the  part  of  plan 
B"  T'  into  any  convenient  number  of  parts,  and  from 
the  points  carry  lines  parallel  to  the  center  line  or  axis 
until  they  cut  the  top  line  u'  V,  and  from  there  extend 
them  in  the  direction  of  Z'  until  they  cut  the  line  B'  C'. 
Place  the  T-square  at  right  angles  to  the  axis,  and, 
bringing  it  against  the  several  points  in  the  line  B'  C', 
which  represents  the  shape  shown  by  E  C  F  in  eleva- 
tion of  side,  cut  the  side  V  X'.  as  shown.  From  Z'  as 
center,  with  Z'  V  as  radius,  describe  the  arc  I'  J',  upon 
which  lay  off  a  stretchout  of  plan.  As  the  part  of  the 
plan  B"  T'  corresponds  to  B'  C',  which  shows  one-half 

of  one  side  of  boat,  and  as 
this  part  of  plan  is  divided 
into  three  parts,  six  of 
these  parts  are  spaced  off 
on  the  arc  I'  J  and  num- 
bered from  1  to  4,  and  4 
to  1,  4  being  the  center 
line.  Through  these  points 
in  the  stretchout  draw 
measuring  lines  to  the  cen- 
ter Z',  as  shown.  AVith 
one  point  of  the  compasses 
set  at  Z',  bring  the  pencil 
point  up  to  the  several 
points  between  V  and  C'  in 
the  elevation,  and  describe 
arcs  cuttingmeasuringlines 
of  corresponding  numbers 

in  the  stretchout;  then  a  line  traced  through  these 
points  of  intersection  will  form  the  line  I'  K'  J',  show- 
ing the  upper  line  of  the  pattern  for  one  side  of  the 
boat. 

To  obtain  the  bottom  line  of  the  pattern,  with  Z' 
as  center  and  radius  Z'  X',  describe  the  arc  M'  N'. 
Divide  the  plan  of  bottom  of  boat,  as  M  T  X  in  Fig. 
488,  into  any  convenient  number  of  equal  parts,  in  this 
case  six,  three  on  each  side  of  the  center  T,  and  start- 
ing from  the  center  line  4  of  pattern,  space  off  three 
spaces  each  way  on  the  arc  M'  N',  thus  establish- 


Pattern  Problems. 


269 


ing   the  points  M'  and  N'  of  pattern,  corresponding 
to    the  points   M    and  N  of  plan.      By  drawing  the 


lines  M'  I'  and  N'  J'  the   pattern  for  one  side  of  the 
boat,  shown  by  E  F  H  G  in  elevation,   is  completed. 


PROBLEM  139. 


Pattern  for  the  Frustum  of  a  Cone  Fitting:  Ag-ainst  a  Surface  of  Two  Inclinations. 


In  Fig.  490,  let  A  B  C  D  represent  the  frustum 
of  a  cone,  the  base  of  which  is  to  be  so  cut  as  to  make 
it  lit  against  a  roof  of  two  inclinations,  as  indicated  by 
P  R  D.  Continue  the  lines  of  the  sides  of  the  cone 
A  B  and  D  C  upward  until  they  meet  in  the  point  X, 
which  is  the  apex  of  the  complete  cone.  Through  the 
apex  of  the  cone  draw  the  line  X  R,  representing  the 
axis  of  the  cone,  meeting  the  ridge  of  the  roof  in  the 
point  R,  and  continuing  downward  in  the  direction  of 
Y,  as  shown.  At  any  convenient  distance  below  A  D 
draw  a  horizontal  line,  G  H,  as  a 
base,  and  immediately  below  it  draw 
a  plan  of  the  same,  as  shown  by  E 
SFY. 

Subdivide  this  plan  into  any 
convenient  number  of  spaces,  as 
indicated  by  the  small  figures  0,  1, 
2,  3,  etc.  From  the  points  thus 
established  carry  lines  vertically 
until  they  cut  the  base  of  the  cone 
G  II,  and  from  this  line  carry  them 
in  the  direction  of  the  apex  X  until 
they  cut  the  line  of  the  given  roof. 
From  the  points  established  in  the 
roof  line  A  R  draw  lines  at  right 
angles  to  the  axis  of  the  cone  X 
Y,  continuing  them  until  they  strike 
the  side  of  the  cone  A  B.  From 
X  as  center,  with  X  G  as  radius,  describe  the  arc  GK, 
upon  which  lay  off  a  stretchout  of  the  plan. 

As  the  pattern  really  consists  of  four  equal  parts 
or  quarters,  the  divisions  of  the  plan  have  been  num- 
bered from  0  to  4  and  from  4  to  0  alternating,  the 
points  0  representing  the  lowest  and  the  points  4  the 
highest  points  of  each  quarter.  Therefore  in  number- 
ing the  points  of  the  stretchout  G  K,  any  point  can  be 
assumed  as  a  beginning  which  is  deemed  the  best  place 
for  the  joint  (in  this  case  4),  numbering  from  4  to  0  and 
reversing  each  time,  all  as  shown.  From  these  points 
established  in  the  arc  G  K  draw  lines  to  the  apex  X. 
Then,  with  X  as  center,  and  with  radii  corresponding  to 


the  points  already  established  in  the  side  B  G  of  the  cone, 
strike  arcs  as  shown  by  the  dotted  lines,  cutting  meas- 


Fig.  490.— Pattern  for  the  Frustum  of  a  Cone  Fitting  Against  a 
Surface  of  Two  Inclinations. 

uring  lines  of  corresponding  number.   Then  a  line  traced 
through  the  points  of  intersection,  as  shown  by  0  L,  will 


270 


The  New  Metal    Worker  Pattern  Book. 


be  the  shape  of  the  pattern  at  the  bottom  and  0  N  M  L 
will  constitute  the  entire  pattern  of  the  frustum  of  a 


cone  adapted  to  set  over  the  ridge  of  u  roof,  as  indi- 
cated in  ihe  elevation. 


PROBLEM    140. 

The  Pattern  of  a  Frustum  of  a  Cone  Intersected  at  its  Lower  End  by  a  Cylinder,  Their  Axes  Intersecting: 

*  at  Right  Angles. 


Let  S  P  K  T  in  Fig.  491  be  the  elevation  of  the 
cylinder,  and  a  G  K  H  b  the  elevation  of  the  frustum. 
Draw  the  axis  of  the  cylinder,  A  B,  which  prolong,  as 
shown  by  C  D,  on  which  construct  a  prolile  of  the 
cylinder,  as  shown  by  C  E  I)  F.  Produce  the  sides  of 
the  frustum,  as  shown  in  the  elevation,  until  they 
meet  in  the  point  L,  which  is  the  apex  of  the  cone. 
Draw  the  axis  L  K,  which  produce  in  the  direction 
of  O,  and  at  any  convenient  point  upon  the  same  con- 
struct a  plan  of  the  frustum  at  its  top,  a  b. 

In  connection  with  the  profile 
of  the  cylinder  draw  a  correspond- 
ing elevation  of  the  cone,  as  shown 
by  K1  a1  b'  K\  Produce  the  sides 
K1  a1  and  K3  b'  until  they  intersect, 
thus  obtaining  the  point  L1,  the 
apex  corresponding  to  L  of  the 
elevation.  Draw  the  axis  L1  E,  as 
shown,  which  produce  in  the  direc- 
tion of  N1,  and  upon  it  draw  a 
second  plan  of  the  frustum  at  a  b, 
as  shown  by  M'  O1  N1.  Divide  the 
plans  M  O  N  and  M1  O1  N'  into  the 
same  number  of  equal  parts,  com- 
mencing at  corresponding  points  in 
each,  as  shown.  With  the  T-square  set  parallel  to 
the  axis  of  the  cones,  and  brought  successively  against 
the  points  in  the  plans,  drop  lines  to  the  lines  a  b  and 
a1  b1,  as  shown. 

From  L1  draw  lines  through  the  points  in  o1  b\  cut- 
ting the  profile  of  the  cylinder,  as  shown  in  K'  E  K2. 
and  in  like  manner  from  the  apex  L  draw  lines  indefi- 
nitely through  the  points  in  a  b.  '  Place  the  T-square 
parallel  to  the  sides  of  the  cylinder,  and,  bringing  it 
against  the  points  in  the  profile  K1  E  K2  just  described, 
cut  corresponding  lines  in  the  elevation,  as  shown  at 
II  K  G.  A  line  traced  through  these  points  of  inter- 
section, as  shown  by  H  K  G,  will  form  the  miter  line 
between  the  two  pieces  as  it  appears  in  elevation. 

This  miter  line  is  not  necessary  in  obtaining  the 
pattern,  but  the  method  of  obtaining  it  is  here  intro- 


duced merely  to  show  how  it  may  be  done,  should  it 
be  desired  under  similar  circumstances  in  any  other 
case.  The  development  of  the  pattern  in  this  case 
could  be  most  easily  accomplished  by  using  L1  as  a 
center  from  which  to  strike  arcs  from  the  various 
points  on  the  line  «'  K'.  The  same  result  is  accom- 


N' 


M 


Fiy.  491. — The  Pattern  of  a  Frustum  of  a  Cone  Intersected  at  Its 
Lower  End  by  a  Cylinder,  Their  Axes  Intersecting  at  Rirjht 
Angles. 

plished,  however,  by  continuing  the  lines  drawn  from 
K1  E  K2  until  they  meet  the  side  a  G  of  the  cone  pro- 
longed, as  shown  from  G  to  Z.  Thus  a  Z  becomes  in 
all  respects  the  same  as  a1  K1. 

From  L  as  center,  and  with  radius  L  a,  describe 
the  arc  //  a*,  upon  which  lay  off  a  stretchout  of  the 
plan  M  0  N  of  the  frustum.  Through  each  of  tin- 
points  in  this  stretchout  draw  lines  indefinitely,  radiat- 
ing from  L,  as  shown,  Number  the  points  in  the 


Pattern  Problems. 


271 


stretchout  «2  b"  corresponding  to  the  numbers  in  the 
profile,  commencing  with  the  point  occurring  where  it 
is  desired  to  have  the  seam.  Set  the  compasses  to 
L  Z  as  radius,  and,  with  L  as  center,  describe  an  arc 
cutting  the  corresponding  lines  drawn  through  the 
stretchout,  as  shown  by  1,  3  and  1.  In  like  manner 


reduce  the  radius  to  the  second  point  in  G  Z,  and  de- 
scribe an  arc  cutting  2,  4,  4  and  2.  Also  bring  the 
pencil  to  the  third  point  and  cut  the  lines  correspond- 
ing to  it  in  the  same  way.  Then  a  line  traced  through 
the  points  thus  obtained,  as  shown  by  H'  K"  G1,  will 
be  the  pattern  of  the  frustum. 


PROBLEM    141. 
The  Pattern  for  a  Conical  Boss. 


The  principles  and  conditions  in  this  problem  arc 
exactly  the  same  as  those  in  the  one  immediately  pre- 
ceding (that  is,  the  frustum  of  a  cone  raitering  against 


Fig.  492.— The  Pattern  fur  a  Conical  Boss. 

a  cylinder,  their  axis  being  at  right  angles),  but  its  pro- 
portions are  so  different  that  it  is  here  introduced  as 
showing  that  the  same  application  of  principles  often 
produces-  results  so  widely  differing  in  appearance  as  to 
be  scarcely  recognizable. 

Let  A  B  C  D  of  Fig.  492  represent  the  elevation 
of  the  boss  that  is  required  to  fit  against  the  cylindrical 


can,  a  portion  of  the  plan  of  which  is  shown  by  the 
arc  A  B.  The  plan  at  the  smaller  end  of  the  boss 
is  represented  by  K  K  G  II.  Continue  the  lines  A  D 
and  B  0  until  they  intersect  at  K, 
which  is  the  apex  of  the  cone  of 
which  the  boss  is  a  frustum.  An  in- 
spection of  the  elevation  will  show  that 
it  is  only  necessary  to  describe  one- 
fourth  of  the  pattern,  the  remaining  parts 
being  duplicates.  Divide  one-quarter  of 
the  plan  into  any  convenient  number  of 
parts,  in  the  present  instance  four,  as 
shown  by  the  points  in  H  E. .  Drop  lines 
from  these  points  to  the  base  D  C,  as 
shown.  Draw  lines  from  K  through  the 
points  in  the  base  until  they  intersect 
the  arc  at  A  B,  which  represents  the 
body  of  the  can.  These  points  can 
be  numbered  to  correspond  with  the 
points  in  the  plan  from  which  they  are 
derived.  At  right  angles  to  the  line  F  K 
draw  lines  from  the  points  on  A  B  until 
they  strike  the  line  A  K,  where  their 
•  true  distances  from  K  can  lie  measured. 
With  K  as  a  center,  and  K  D  as  radius, 
strike  the  arc  L  M  N,  equal  in  length  to 
the  circumference  of  plan. 

If  the  whole  pattern  of  boss  is  to  be 
described    from    measurements    derived 
from    elevation   it  will   be    necessary   to 
reverse  the  order  of  the  numbers  for  each 
quarter,  as  shown.      From  K  draw  lines 
extending     outwardly      through      these 
points,  as  indicated  by  the  small  figures.      With  K  as 
center,  draw   an  arc  from  the   point   1'  until  it    inter- 
sects radial   lines  1  drawn  from  K,  as   shown  at  0,  Z 
and  R.      In  the  same  manner  draw  an   arc   from  2'   to 
lines  '2,  &c.,  as  shown.     A  line  traced  through  these 
points    will  produce  the  desired  patterns,  as  shown  by 
L  0  S  Z  V  R  N. 


272 


The  \>  a-  .\fctal    Worker  Pattern  Book. 


PROBLEM    142. 

Pattern  for  the  Lip  of  a  Sheet  Metal  Pitcher. 


Let  A  B  C  D  of  Fig.  493  represent  the  side  ele- 
vation of  a  pitcher  top  having  the  same  flare  all  around, 
iind  E  F  G  H  the  plan  at  the  base.  By  producing  the 
lines  A  D  and  B  C  until  they  intersect  in  the  point  K, 
the  apex  of  the  cone  of  which  the  pitcher  top  is  a  sec- 
tion will  be  obtained.  Divide  one-half  of  the  plan  into 
any  convenient  number  of  equal  spaces,  as  shown  by 
the  points  in  G  H  E.  From  these  points  drop  lines  lo 
tin1  base  D  C,  as  indicated.  Then  draw  radial  lines 
from  K,  cutting  the  points  in  D  C,  and  producing 
them  until  they  intersect  the  curved  line  representing 
the  top  of  the  pitcher,  otherwise  an  irregular  cut 
through  the  cone,  as  shown  by  A  B.  For  convenience 
in  subsequent  operations,  number  these  points  to  cor- 
respond with  the  numbering  of  the  points  in  the  plan 
from  which  they  are  derived. 

Place  the  T-square  at  right  angles  to  the  axial  line 
of  the  cone  II  K,  and,  bringing  it  against  the  several 
points  in  A  B,  cut  the  line  DA,  as  shown  in  the 
diagram.  By  this  means  there  will  be  obtained  in  the 
line  K  A  the'  length  of  radii  which  will  describe  area 
corresponding  to  points  in  the  top  line  of  the  lip  A  B. 
With  K  as  center  and  K  D  as  radius,  describe  the  arc 
L  M  N,  which  in  length  make  equal  to  the  circumfer- 
ence of  the  plan  by  stepping  off  on  it  spaces  equal  to 
the  spaces  originally  established  in  the  plan,  all  as  in- 
dicated by  the  small  figures.  From  K  through  each 
of  the  points  in  L  M  N  thus  established  draw  radial 
lines,  extending  outwardly  indefinitely,  as  shown. 
Then  with  K  as  center  and  K",  K,'  K3,  etc.,  as  radii, 
strike  arcs,  which  produce  until  they  intersect  radial 
lines  of  corresponding  number  just  drawn,  all  as  shown 
in  the  diagram.  Then  a  line  traced  through  the  points 
thus  obtained  will  be  the  required  pattern,  all  as  shown 
by  L  0  P  R  N  M. 

The  method  above  described  is  a  strictly  mathe- 
matical rule  for  obtaining  such  shapes  when  a  design 
embodying  the  necessary  curve  at  the  top  is  at  hand. 
As  by  the  nature  of  the  problem,  this  part  of  the  pat- 
tern does  not  require  to  be  fitted  or  joined  to  any 
other  piece,  it  would  be  much  easier  to  obtain,  by  the 
foregoing  method,  the  principal  points  in  the  outside 
curve  of  the  pattern  and  finish  by  drawing  the  re- 
mainder to  suit  the  taste  of  the  designer.  In  other 
words,  after  the  arc  L  M  N  has  been  drawn  and 
stepped  off  into  spaces,  draw  radial  lines  from  K 
through  the  points  representing  the  highest  and  the 


lowest  parts  required  in  the  top  curve,  as  (»."»  and  12, 
"upon  which  lines  the  required  lengths  can  be  set  <>IT. 
Then  these  points  can  be  connected  by  anv  curve 
suitable  for  the  purpose. 

The  principle  involved  in  the  foregoing  is  exactlv 
the  same  as  that  of  a  hip  or  sitz  bath  given  in  the  fol- 


\ 


Fig,  493,— Pattern  for  the  L<p  of  a  Sheet  Metal  Pitcher. 

lew  ing  problem,  the  difference  in  the  finished  article 
being  a  matter  of  size  and  shape  and  not  of  principle. 
Thus  the  sides  could  be  made  less  flaring  by  placing 
the  point  K  much  further  away  from  the  base  line,  the 
hight  A  D  could  be  increased  and  the  curve  A  B 
could  be  altered  to  one  more  desirable;  but  the  various 
steps  necessary  to  perform  the  task  would  remaiii 
exactly  the  same. 


Pattern 

PROBLEM  143. 

Pattern  for  a  Hip  Bath  of  Regular  Flare. 


273 


Let  C  B  ].)  K.  in  Fig-  404-,  IK-  the  elevation  <>f  the 

lv  <>[  ;t  hip  bath  having  an  equal  amount  of  llaiv  on 
all  .sides,  the  plan  of  which   is  a  circle.      In   describing 


At  any  convenient  distance  above  D  draw  J  K 
parallel  to  C  E  to  be  used  as  a  regular  base  upon  which 
to  measure  the  circumference  of  the  cone.  Parallel  to 
.1  K  draw  F  II,  and  from  a  center  obtained  on  F  II  by 
prolonging  the  axis  A  X  draw  a  half -plan  of  the  frustum, 
as  shown  by  F  G  II.  Divide  this  half-plan  into  any 
convenient  number  of  equal  parts,  and  from  the  points 
thus  obtained  carry  lines  parallel  to  the  axis  until  they 
cut  the  line  J  K,  and  from  there  extend  them  in  the 
direction  of  the  apex  A,  thus  cutting  the  curved  line 
B  D.  Place  the  T-square  parallel  with  J  K,  and  bring- 
ing it  against  the  several  points  in  the  curved  line  B  D, 
cut  the  side  E  D,  as  shown.  From  A  as  center,  with 


K2 


Fig.  494.—  Pattern  for  a  Hip  Bath  of  Regular  Flare, 


the  pattern  for  the  body  it  will  be  considered  as  a  sec- 
tion of  a.  right  cone,  the  plane  C  E  being  at  right 
angles  to  the  axis  and  the  base  being  represented  hy 
the  curved  line  B  D,  as  shown.  The  sides  E  D  and 
C  B  can  be  extended  until  they  meet  at  A.  Then  A 
will  be  the  apex  of  a  cone  of  which  C  B  D  E  is  a  frus 
turn  having  an  irregular  base  B  D. 


A  K  as  radius,  describe  the  arc  K'  K2,  on  which  lay  off 
a  stretchout  of  either  one-half  or  the  whole  of  the  plan, 
as  may  be  desired.  In  this  case  a  half  is  shown.  From 
the  extremities  of  this  stretchout,  as  K'  and  K2,  draw 
lines  to  the  center,  as  K'  A  and  K"  A,  and  from  the 
several  points  in  the  stretchout  draw  similar  lines,  as 
shown  by  1,  2,  etc.  With  one  point  of  the  dividers 


The  New  Metal   Worker  Pattern  Book. 


set  at  A  bring  the  pencil  point  to  the  point  D  in  the 
side  A  l\.  ami  with  that  radius  describe  an  arc,  which 
produce  until  it  cuts  the  corresponding  line  1'2  in  the 
stretchout,  as  shown  at  D1.  In  like  manner,  bringing 
the  pencil  point  up  to  the  several  points  between  D 
and  E  in  the  elevation,  describe  arcs  cutting  lines  of 
corresponding  numbers  in  the  stretchout.  Then  a  line 
traced  through  these  intersections  will  form  the  upper 
line  of  the  pattern.  From  A  as  center,  with  A  E  as 
radius,  describe  the  arc  C'  E1,  cutting  A  K1  and  A 
K",  as  shown  by  C1  and  E1,  forming  the  lower  line  of 
pattern.  Then  C1  Bl  D1  E1  will  be  half  the  pattern 
for  the  side  of  the  hip  bath. 


As  a  feature  <>!'  design,  the  form  produced  in  the 
pattern  by  a  curved  line  B  D  drawn  arbitrarily  may 
not  be  entirely  satisfactory.  If,  for  instance,  that  part, 
of  the  pattern  lying  between  lines  9  and  12  should  not 
appear  as  desired,  it  can  be  moditied  upon  the  pattern 
at  will,  as  this  edge  of  the  pattern  is  not  required  to  lit 
any  other  form.  Such  a  modification  is  shown  bv  the 
dotted  lines  </'  lv3  of  the  pattern  and  a  K  of  the  eleva- 
tion. The  Coot  of  the  tub  is  a  simple  frustum  of  a 
right  cone,  the  pattern  for  which  is  obtained  in  the 
manner  described  in  Problem  123.  Different  forms  of 
bathtubs  in  which  the  flare  is  irregular  will  be  found 
in  Section  3  of  this  chapter. 


PROBLEM   144. 
The  Envelope  of  a  Frustum  of  a  Right  Cone  Contained  Between  Planes  Oblique  to  Its  Axis. 

In  Fig.  4y.5,  let  F  L  M  K  represent  the  section  of 
the  cone  the  pattern  for  which  is  required.  Produce 
the  sides  F  L  and  K  M  until  they  meet  in  the  point  N, 
which  is  the  apex  of  the  cone  of  which  F  L  M  K  is  a 
frustum.  Through  N  draw  N  E.  bisecting  the  angle 
L  X  M  and  constituting  the  axis  of  the  cone,  which  pro- 
duce in  the  direction  of  D  indefinitely.  From  K  draw 
K  II  at  right  angles  to  the  axis.  At  any  convenient 
distance  above  the  cone  construct  a  plan  or  profile  as  it 
would  appear  when  cut  on  the  line  K  II.  letting  the 
center  of  the  profile  fail  upon  the  axis  produced,  all  as 
shown  by  A  1)  C  B.  Divide  the  profile  into  any  num- 
ber of  equal  parts,  and  from  the  points  thus  obtained 
draw  lines  parallel  to  the  axis,  cutting  K  II.  From 
the  apex  X,  through  the  points  in  K  II,  draw  lines 
cutting  the  top  L  M  and  the  base  F  K.  Place  the 
blade  of  the  T-square  at  right  angles  to  the  axis  of  the 
cone,  and,  bringing  it  successively  against  the  points 
in  L  M  and  F  K,  cut  the  side  N  F,  as  shown  above  L, 
and  from  H  to  F.  From  N  as  center,  with  radius 
N  II,  strike  the  arc  T  S  indefinitely,  upon  which  lay  off 
a  stretchout  from  the  plan,  as  shown,  and  through 
the  points  of  which,  from  the  center  N",  draw  lines  in- 
definitely, as  shown.  With  the  point  of  the  compasses 
still  at  N,  and  the  pencil  brought  successively  against 
the  points  in  the  side  from  H  to  F,  describe  arcs, 
which  produce  until  they  cut  corresponding  lines  drawn 
through  the  stretchout.  Then  a  line  traced  through  Fl'J'  495-—The  En™l°Pe  of  a  Frustum  of  a  Right  Cone  Contained 

.  .    .  between  Planes  Oblique  to  its  Axis. 

these  points  of  intersection,  as  shown  by  T  TJ  S,  will 

form  the  lower  line  of  pattern.     In  like  manner  draw  ing  lines  drawn  through  the  stretchout.     A  line  traced 

arcs  by  radii  corresponding  to  the  points  in  the  side  at  through  these  points,  as  R  P  0,  will  be  the  upper  line 

L,  which  produce  also  until  they  intersect  correspond-  of  the  pattern  sought. 


Pattern  Problems. 


275 


PROBLEM    145. 


The  Pattern  of  a  Cone  Intersected  by  a  Cylinder  at  Its  Upper  End,  Their  Axes  Crossing:  at  Right  Angles. 


In  the  plan,  Fig.  -IDfi,  let  A  B  C  D  F  represent  a 
frustum  of  the  cone  B  G  G,  B  H  C  being  the 
half  profile  of  cone  at  its  base  and  A  D  J  the 
plan  of  the  cylinder.  In  line  with  the  cylinder  in  plan 
draw  the  elevation,  as  shown  by  K  S  T  U.  With  the 
J-square  placed  parallel  to  the  sides  of  the  cylinder, 
carry  a  line  from,  the  point  G  in  plan  to  any  convenient 
point,  as  G'  of  elevation.  At  right  angles  to  G  G' 
draw  G'  Q  indefinitely,  and  extend  15  C  through  Q 
G',  cutting  same  at  O.  With  ()  as  center,  and  E  C 
of  plan  as  radius,  describe  the  semicircle  L  Q  M,  rep- 
resenting one-half  of  the  profile  of  the  cone  at  the 
larger  end.  Divide  L  Q  M  into  any  convenient  num- 
ber of  equal  parts,  as  indicated  by  the  small  figures. 
From  the  points  thus  obtained  carry  lines  at  right 
angles  to  L  M,  cutting  that  line  as  indicated.  From 
the  points  thus  obtained  in  L  M  carry  lines  to  the  apex 
G',  as  indicated  by  the  dotted  lines  in  the  engraving. 
Divide  B  H  C  into  the  same  number  of  equal  parts  as 
was  L  Q  M,  numbering  them  to  correspond  with  the 
elevation,  as  shown,  and  from  the  points  thus  obtained 
carry  lines  at  right  angles  to  B  C,  cutting  that  line  as 
indicated.  From  the  points  in  B  C  carry  lines  to  the 
apex  G,  cutting  the  plan  of  cylinder  as  shown.  With 
the  blade  of  the  T-square  placed  parallel  with  G  G', 
and  brought  successively  against  the  points  thus  es- 
tablished in  'the  plan  of  the  cylinder,  cut  lines  of  corre- 
sponding number  drawn  from  the  points  in  L  M  to 
th 3  apex  G',  as  indicated  from  K  to  N,  and  extend 
th_\so  lines  to  the  line  M  G'.  A  lino  traced  through 
th;sc  points  of  intersection,  as  shown  by  KPN,  rep- 
resents the  intersection  of  cone  with  cylinder  in  eleva- 
'ion,  as  shown  by  A  F  D  in  plan. 

For  the  pattern  proceed  as  follows :  From  G'  as 
•enter,  with  G'  M  as  radius,  describe  an  arc,  as  indi- 
cated by  /  o  m,  and,  starting  from  I,  step  off  the  stretch- 
out of  the  half  profile  L  Q  M,  as  indicated  by  the 
small  figures.  If  the  entire  pattern  is  required  in  one 
piece  extend  the  arc  I  o  m,  and  from  m  set  off  a  du- 
plicate of  I  o  m,  numbering  the  points  in  inverse 
order.  From  the  points  thus  obtained  draw  radial 
lines  to  G',  as  indicated.  .Then  with  G'  as  center,  and 
with  radii  corresponding  to  the  distance  from  G'  to  the 
points  established  in  M  G',  describe  arcs,  producing 


them  until  they  cut  lines  of  corresponding  number 
drawn  from  G'.  A  line  traced  through  these  points  of 
intersection,  as  shown  by  k  p  n,  will  with  I  o  m  give 


Fig.  496- — Pattern  of  a  Cone  Intersected  &;/  a  Cylinder  at  its  Upper  End. 

the  pattern  of  part  of  article  shown  in  elevation  by 
L  M  N  P  K. 

It  will  be  easily  seen  that  the  pattern  might  have 


276 


The  New  Metal   Worker  Pattern  Book. 


been  obtained  directly  from  the  plan  without  the 
trouble  of  drawing  the  elevation,  as  in  Problems  141, 
142  and  143.  Should  it  be  desirable,  however,  to  cut 
an  opening  in  the  side  of  the  cylinder  to  fit  the  frus- 
tum of  the  cone,  the  bights  of  all  points  in  the  perim- 


eter  of  such  opening  must  be  obtained  from  the  line 
N  P  K  of  the  elevation,  while  width  of  the  opening 
upon  lines  corresponding  to  these  points  must  b<> 
measured  from  F  toward  D  or  A  unou  the  circumfer- 
ence of  the  cylinder. 


PROBLEM    146. 

Pattern  of  a  Tapering;  Article  with  Equal  Flare  Throughout,  which  Corresponds  to  the  Frustum  of  a 
Cone  Whose  Base  Is  an  Approximate  Ellipse  Struck  from  Centers,  the  Upper  Plane  of 

the  Frustum  Being-  Oblique  to  the  Axis. 


In  Fig.  497,  let  H  F  G-  A  be  the  shape  of  -the 
article  as  seen  in  side  elevation.  The  plan  is  shown 
by  I  L  N  O.  In  order  to  indicate  the  principle  in- 
volved in  the  development  of  this  shape,  it  will  be 
necessary  first  to  analyze  the  figure  and  ascertain  the 
shape  of  the  solid  of  which  this  frustum  is  a  part. 
Since  by  the  conditions  of  the  problem  the  base  is 
drawn  from  centers  and  the  sides  have  equal  flare,  it 
follows  that  each  arc  used  in  the  plan  of  the  base  is  a 
part  of  the  base  of  a  complete  cone -whose  diameter  can 
be  found  by  completing  the  circle  and  whose  altitude 
can  be  found  by  continuing  the  slant  of  its  sides  till 
they  meet  at  the  apex,  all  of  which  can  be  seen  by  an 
inspection  of  the  engraving.  Thus  those  parts  of  the 
figure  shown  in  plan  by  K  U  T  M  and  R  U  T  P  may 
be  considered  as  segments  cut  from  a  right  cone,  the 
radius  of  whose  base  is  either  0  K  or  L  R,  and 
whose  apex  E  is  to  be  ascertained  by  continuing  the 
alant  of  the  side  LJ  C1  till  it  meets  a  vertical  line 
erected  from  O1  of  the  plan,  which  is  the  center  of  the 
arc  of  the  base,  all  as  shown  in  the  end  view.  Also 
those  parts  of  the  plan  shown  by  K  U  R  and  M  T  P 
are  segments  of  a  right  cone  whose  radius  is  U  I  or  T 
N"  and  whose  altitude  is  found,  as  in  the  previous  case, 
by  continuing  the  slant  of  its  side  G.A  (which  is  par- 
allel to  C1  I/)  till  it  meets  a  vertical  line  erected  from 
its  center  T,  as  shown  in  the  side  view. 

To  complete  the  solid,  then,  of  which  FGA  H  is 
a  frustum,  it  will  only  be  necessary  to  take  such  parts 
of  the  complete  cones  just  described  as  are  included 
between  the  lines  of  the  plan  and  place  them  together, 
each  in  its  proper  place  upon  the  plan.  The  resulting 
figure  would  then  have  the  appearance  shown  by  H  D 
C  B  A  when  seen  from  the  side,  and  that  of  0s  C1  L2 


when  seen  from  the  end.  The  lines  of  projection  con- 
necting the  various  views  together  with  the  similarity 
of  letters  used  will  show  the  correspondence  of  parts. 
This  figure  is  made  use  of  in  the  second  part  of 
Chapter  V,  Principles  of  Pattern  Cutting,  to  which 
the  reader  is  referred  for  a  further  explanation  of  prin- 
ciples. 

Divide  one-half  of  the  plan  into  any  convenient 
number  of  equal  parts,  as  shown  by  the  small  figures, 
and  from  the  points  thus  established  carry  lines  verti- 
cally, cutting  the  base  line  H  A,  and  thence  carry  them 
toward  the  apexes  of  the  various  cones  from  the  bases 
of  which  they  are  derived.  That  is,  from  the  points 
upon  the  base  line  H  A  derived  from  the  arc  K  M 
draw  lines  toward  the  apex  E,  and  from  the  points  de- 
rived from  the  arc  I  K  carry  lines  toward  the  apex  D, 
and  in  like  manner  from  the  points  derived  from  the 
arc  M  N  carry  lines  in  the  direction  of  the  apex  B,  all 
of  which  produce  until  they  cut  the  top  line  F  G  of 
the  article.  From  the  points  in  F  G  thus  established 
carry  lines  to  the  right,  cutting  the  slant  lines  of  the 
cones  to  which  they  correspond.  Tims,  from  the 
points  occurring  between  F  and  /  draw  lines  cutting 
B  A,  being  the  slant  of  the  small  cone,  as  shown  bv 
the  points  immediately  below  W.  In  like  manner, 
from  the  points  between  g  and  G  carry  lines  cutting 
the  same  line,  as  shown  at  G.  The  slant  line  of  the 
large  cone  is  shown  only  in  end  elevation,  and  there- 
fore the  lines  corresponding  to  the  points  between  / 
and  g  must  be  carried  across  until  they  meet  the  line 
B1  L1. 

Commence  the  pattern  by  taking  any  convenient 
point,  as  E1,  for  center,  and  E'  La  as  radius,  and  strike 
the  arc  L3  S  indefinitely.  Upon  this  arc,  commencing 


Pattern  Problems. 


277 


at  any  convenient  point,  as  K4,  set  off  that  part  of  the 
stretchout  of  tin-  plan  corresponding  to  the  base  of  the, 
larger  cone,  as  shown  by  the  points  o  to  13  in  the  plan, 
ami  as  indicated  by  corresponding  points  from  K*  to 
M2  in  the  arc.  From  the  points  thus  established  draw 
lines  indefinitely  in  the  direction  of  the  center  E1,  as 


aa  shown  by/1  g\  Next  take  A  B  of  the  side  eleva- 
lion  as  radius,  and,  setting  one  fool  (>f  (lie  compasses  in 
the  point  K'  of  the  arc,  establish  the  point  1)'  in  the 
line  K'  K1,  and  in  like  manner,  from  M'J,  with  the  saijie 
radius,  establish-  the  point  B2  in  the  line  M"  K',  which 
will  be  the  centers  from  which  to  describe,  those  parts 
of  the  patterns  derived  from  the  smaller  cones.  From 
I)1  and  IV  as  centers,  with  radius  B  A,  strike  aroe  from 
K'  and  M2,  respectively,  as  shown  by  K'  F  and  M'  N1, 
upon  which  set  off  those  parts  of  the  stretchout  cor- 
responding to  the  smaller  cones,  as  shown  b\-  the  arcs 
K  I  and  M  X  of  the  plan.  From  the  points  thus  estab- 
lished, being  5  to  1  and  13  to  17,  inclusive,  draw  radial 
lines  to  the  centers  D'  and  BJ,  as  shown. 


-Hif  / 

IV  /  /  \  \ 


/--/--I — 

— -J —  -l     l    — j — 


fiij.  497.—  Pattern  of  the  Frustum  of  a  Cone,  the  Base  of  which  is 
an  Approximate.  Ellipse  Struck  from  Centers,  the  Upper  Plane  of 
tlte  frustum  being  Oblique  to  the  Axis. 

shown.  From  E'  as  center,  with  radii  corresponding 
to  the  distance  from  E1  to  points  5  to  13  inclusive, 
established  in  the  line  B1  L5  already  described,  cut  cor- 
responding radial  lines  just  drawn,  and  through  the 
points  of  intersection  thus  established  draw  a  line,  all 


For  that  part  of  the  pattern  shown  from  F'  to  /', 
set  the  dividers  to  radii,  measuring  from  B,  corre- 
sponding to  .the  several  points  immediately  below  W  of 
the  side  elevation,  and  from  D'  as  center  cut  the  cor- 
responding radial  lines  drawn  from  the  arc.  In  like 
manner,  for  that  part  of  the  pattern  shown  from  G1  to 
g',  set  the  dividers  to  radii  measured  from  B,  corre- 
sponding to  the  points  in  the  line  B  A  at  G,  with  which, 
from  B"  as  center,  strike  arcs  cutting  the  correspond- 
ing measuring  lines,  as  shown.  Then  F'  G1  N1  F  will 
be  one-half  of  the  pattern  sought — -in  other  words,  the 
part  corresponding  to  I  K  L  M  N  of  the  plan.  The 
whole  pattern  may  be  completed  by  adding  to  it  a  du- 
plicate of  itself. 


278 


The  New  Metal   Worker  Pattern  Book. 


PROBLEM    147. 
The  Envelope  of  a  Right  Cone,  Cut  by  a  Plane  Parallel  to  Its  Axis. 


Let  B  A  F  in  Fig.  498  be  a  right  cone,  from  which 
a  section  is  to  be  cut,  as  shown  by  the  line  C  1)  in  the 
elevation.  Let  G  L  H  K  be  the  plan  of  the  cone  in 
which  the  line  of  the  cut  is  shown  by  D'  D".  For 
the  pattern  proceed  as  follows:  Divide  that  portion 
of  the  plan  corresponding  to  the  section  to  be  cut  off, 
as  shown  by  D1  G  D",  into  as  many  spaces  as  are  nec- 
essary to  give  accuracy  to  the  pattern,  and  divide  the 
remainder  of  the  plan  into  spaces  convenient  for  laying 
off  the  stretchout.  From  A  as  center,  with  radius  A 
B,  describe  an  arc,  as  M  N,  which  make  equal  to  the 
stretchout  of  the  plan  G  L  H  K,  dividing  it  into  the 
same  spaces  as  employed  in  the  plan,  taking  care  that 
its  middle  portion,  D"  D',  is  divided  to  correspond  with 
D'D'of  the  plan.  From  the  points  in  M  N  correspond- 
ing to  that  portion  of  the  plan  indicated  by  D1  G  D'— 
namely,  8  to  16  inclusive — draw  lines  to  the  center  A, 
as  shown. 

From  points  of  the  same  number  in  the  plan  carry 
lines  vertically,  cutting  the  base  of  the  cone,  as  shown 
from  B  to  D,  and  thence  continue  them  toward  the  apex 
A,  cutting  C  D,  as  shown.  From  the  points  in  C  D 
carry  lines  at  right  angles  to  the  axis  A  E  cutting  the 
side  of  the  cone,  as  shown  by  the  points  between  C 
and  B.  From  A  as  center,  with  radii  corresponding  to 
the  distances  from  A  to  the  several  points  between  C 
and  B,  cut  lines  drawn  from  points  of  corresponding 
number  in  the  stretchout,  to  A,  and  through  the 
points  of  intersection  thus  obtained  trace  a  line,  as 
shown  by  D3  (7  D4.  Then  the  space  indicated  by  D3 
C2  D4  is  the  shape  to  be  cut  from  the  envelope  M  A  IS" 
of  the  cone  to  produce  the  shape  to  fit  against  the  line 
C  D  in  the  elevation. 

To  obtain  the  pattern  of  a  piece  necessary  to  fill 
the  opening  D3  C"  D*  in  the  envelope,  and  represented 
by  C  D  of  the  elevation,  draw  any  vertical  line,  through 
•which  draw  a  number  of  horizontal  lines  corresponding 
in  hight  to  the  points  in  C  D.  The  width  of  the  piece 
upon  each  of  these  lines  may  be  found  by  measuring 


7      e 
K 

Fig.  498. -The  Envelope  of  a  Right   Cone   Cut  by  a  Plane 
Parallel  to  its  Axis. 

across  the  plan  upon  lines  of  corresponding  number,  as 
11  13,  10  14,  etc.  Such  a  section  is  properly  called  a 
hyperbola  (see  Def.  113,  Chap.  I). 


PROBLEM    148. 
The  Pattern  for  a  Scale  Scoop,  Having  Both  Ends  Alike. 


In  Fig.  499,  let  A  B  C  D  represent  the  side  ele- 
vation of  a  scale  scoop, -of  a  style  in  quite  general 


use,    and  E  F  II  G  a  section  of  the  same  as  it  would 
appear  cut  upon   the   line  B  D,  or,  what  is  the  same, 


Pattern  Problems. 


so  far  as  concerns  the  development  of  the  patterns, 
an  end  elevation  of  the  scoop.  The  curved  line  ABC, 
representing  the  top  of  the  article,  may  be  drawn  at 
will,  being,  in  this  case,  a  free-hand  curve.  For  the 
patterns  proceed  as  follows :  From  the  center  K,  by 
which  the  profile  of  the  section  or  end  elevation  is 


Fiij.  Vi'.i.~T.\e  Pattern  for  a  Scale  Scoop. 

drawn,  draw  a  horizontal  line,  which  produce  until  it 
meets  the  center  line  of  the  scoop  in  the  point  0. 
Produce  the  line  of  the  side  D  C  until  it  meets  the  line 
just  drawn  in  the  point  X.  Then  X  is  the  apex  and 


X  0  the  axis  of  a  cone,  a  portion  of  the  envelope  of 
which  each  half  of  the  scoop  may  be  supposed  to  be. 

Divide  one -half  of  the  profile,  as  shown  in  end 
elevation  by  E  G,   into    any    convenient   number   of 
spaces,  and  from  the  points  thus  obtained  carry  lines 
horizontally,   cutting   the  line    B  D,   as    shown,   and 
thence  carry  lines  to  the  point  X, 
cutting    the   top    B    C,    as   shown. 
With  X  D  as  radius,  and  from  X 
as  center,  describe  an  arc,  as  shown 
by  L  N,  upon  which   lay   off   the 
stretchout  of  the  scoop,  as    shown 
in  end  elevation.     From  the  points 
in  L  N  thus  obtained   draw   lines 
to  the    center  X,  as  shown.     From 
the  points  in  B  C  drop  lines  at  right 
angles  to  0  X,  cutting  the  side  D  C, 
as  shown.     "With  X  as  center,  and 
radii  corresponding  to  each   of   the 
several    points    between   D  and  C, 
describe  arcs,  which  produce  until 
they  cut  radial  lines  of  correspond- 
ing   numbers    drawn    from    points 
in  the  arc  L  N  to  the  center  X.     Then  a  line  traced 
through  the  points  thus  obtained,  as  shown  by  L  M  N, 
will  be  the  profile  of  the  pattern  of  one-half  of  the  re- 
quired article. 


PROBLEM    149. 


The  Patterns  for  a  Scale  Scoop,  One  End  of  Which  Is  Funnel  Shaped. 


In  Fig.  500  is  shown  a  side  view  of  a  scale  scoop 
by  which  it  will  be  seen  that  the  portion  A  B  G  II  of 
the  funnel-shaped  end  is  a  simple  cylinder  and,  there- 
fore, need  not  be  further  noticed  hen'.  In  Fig.  501 
are  shown  a  side  and  an  end  elevation  of  the  tapering 
portions.  It  will  also  be  seen  that  the  part  D  E  F  of 
the  side  view  is  similar  in  all  respects  to  the  article 
treated  in  the  preceding  problem,  and  the  pattern 
shown  in  connection  with  the  same  is  obtained  by 
exactly  the  same  method  as  that  there  described  and 
need  not,  therefore,  be  repeated. 

An  inspection  of  the  side  elevation  will  show  that 
the  part  G  B  C  D  F  is  a  section  of  a  cone  of  which  I 
is  the  apex,  H  F  the  base  and  II'  C"  F'  the  plan  of 
the  base,  and  that  this  cone  is  cut  by  the  lines  B  G 
and  C  D.  To  obtain  the  pattern  of  this  part,  first 


divide  F'  C"  and  C"  H'  of  end  elevation  into  any  con- 
venient number  of  parts,  and  from  the  points  thus  ob- 


Fig.  500.— Scale  Scoop,  One  End  of  which  is  Funnel  Shaped. 

tained    carry   lines    cutting  the   miter   line    H  F,    as 
shown.     From  the  points  in  H  F  carry  lines  to  the 


280 


Tlie  New  Metal   ll'//-/,-e/-  Pattern  Boole, 


apex  I,  cutting  the  curved  line  C  D,  as  shown.      From 
the  points  in  C  D  drop  perpendiculars  cutting  the  sides 


the  same  as  in  Fig.  ."><>1.     With  I  of  Pig.  502  as  cen- 
ter,  and  IG  and  1  F  as   radii,  describe   the  arcs  It  8 


/ 


END  ELEVATION 

Fig.  501. — Side  and  End  View  of  Conical  Portion  of  Scoop, 
with  Pattern  of  Piece  D  E  F. 


PATTERN 


2         1         * 
Fig.  BOS.  -Pattern  nf  Piece  B  C  D  F  G  of  Fig.  600. 


G  F,  as  shown.  For  convenience  in  describing  the 
pattern  a  duplicate  of  the  side  view  of  this  part  is 
shown  in  Fig.  502,  in  which  similar  parts  are  lettered 


and  P  Q.  Upon  P  Q  lay  out  twice 
the  stretchout  of  H'  C"  F'  of  Fig. 
501,  if  the  pattern  is  desired  in  one 
piece.  Thus  the  stretchout  of  II'  ('" 
is  represented  l>v  P  ('  and  (^  \V, 
from  the  points  in  which  draw  lines 
to  the  center  I.  From  I  as  center, 
and  radii  corresponding  to  the  dis- 
tance from  I  to  each  of  the  various 
points  in  G  F,  descrilie  ;nvs  cutting 
lines  of  similar  numbers.  Trace  lines 
through  the  points  thus  obtained, 
as  shown  by  T  U,  V  W.  Tin-  H  T  U  W  V  S  is  the 
pattern  for  part  of  scoop  shown  in  side  elevation  l>v 
BCD 


PROBLEM    150. 


The  Pattern  of  a  Conical  Spire  Mitering  upon  Four  Gables. 


Let  E  I  0  B  in  Fig.  503  be  the  elevation  of  a  pin- 
nacle having  four  equal  gables,  down  upon  which  a 
conical  spire  is  required  to  be  mitered,  as  shown. 


Produce  the  sides  of  the  spire  until  they  meet  in  the 
apex  D.  Also  continue  the  side  E  F  downward  to  anv 
convenient  point  below  the  junction  between  the  spire 


Pattern  Problems. 


281 


and  the  gables,  as  shown  by  II,  which  point  may  be 
considered  the  base  of  a  cone  of  which  the  spire  is  a 
part.  Let  A"  K  L  M  be  the  plan  of  the  gables^  The 
diagonal  lines  V  L  and  M  K  represent  the  angles  or 
valleys  between  the  gables,  while  R  S  and  T  U  repre- 
sent the  ridges  of  the  gables  over  which  the  spire  is  to 
be  fitted.  Through  the  point  II  in  the  elevation  draw 
a  line  to  the  center  of  the  cone  and  at  right  angles  to 
its  axis,  as  shown  by  II  C.  -This  will  represent  the 
half  diameter  or  radius  of  the  cone  at  its  base.  With 
radius  C  H,  and  from  center  A"  of  the  plan,  describe  a 
circle,  as"  shown,  which  will  represent  the  plan  of  the 
cone  at  its  base. 

At  any  convenient  distance  from  the  elevation, 
and  to  one  side,  project  a  diagonal  section  correspond- 
ing to  the  line  M  A"  in  the  plan,  as  follows:  From 
all  the  points  in  the  side  of  the  pinnacle  draw  horizon- 
tal lines  indefinitely  to  the  left,  which  will  establish 
the  hights  of  the  corresponding  points  in  the  section. 
Fro  in  any  vertical  line,  as  D'  A',  as  a  center'line  set  off 
upon  the  horizontal  lines  the  distances  as  measured 
upon  the  line  M  A"  of  the  plan.  Thus  make  B'  A1 


5    433 


F2 
1 

Z 
3 
4 


of  the  crossing  of  the  two  ridges  of  the  gables,  there- 
fore a  line  drawn  from  F1  to  B1  will  represent  one  of 
the  valleys  between  the  gables.  Draw  H'  D1,  the  side 
of  the  cone.  Its  intersection  with  the  line  of  the  val- 
ley at  G  will  then  represent  the  hight  of  the  lowest 
points  of  the  spire  between  the  gables,  and  a  line  pro- 


D1 


I. 


Fig.  504.-Pattern. 


M  U 

Fig.  5na— Plan,  Elevation  and  Diagonal  Section. 


The  Pattern  of  a  Conical  Spire  Mitering  Upon  Four  Gables. 

equal  to  M  A"  and  C'  IP  equal  to  A'  5,  the  radius  of      jected  from  this  point  back  into  the  elevation,  as  shown, 
the  cone  at  its  base.     The  point  F'  represents  the  hight      will  locate  those  points  in  that  view. 


Tli.e  New  Metal    Worker  Pattern  Boole. 


To  describe  the  pattern,  first  divide  one-eighth  of 
the  plan  of  the  cone,  choosing  the  one  which  miters 
with  the  gable  shown  in  the  elevation,  into  any  num- 
ber of  equal  spaces,  as  shown  by  the  small  figures. 
From  these  points  carry  lines  vertically  cutting  the 
base  of  the  cone  H  C,  as  shown,  and  thence  toward  the 
apex  D,  cutting  the  lineB  J  of  the  gable,  against  which 
this  part  of  the  cone  is  to  miter.  As  the  true  distance 
of  any  one  of  the  points  just  obtained  upon  the  line  B 
J  from  the  apex  D  can  only  be  measured  on  a  drawing 
when  that  point  is  shown  in  profile,  proceed  to  drop 
these  points  horizontally  to  the  profile  line  D  H,  where 
they  are  marked  I1,  21,  etc.,  and  where  their  distances 
from  D  can  be  measured  accurately.  Next  draw  any 
straight  line,  as  D'  H"  of  Fig.  504,  upon  which  set  off 
all  the  distances  upon  the  line  D  H  of  the  elevation, 
all  as  shown,  each  point  being  lettered  or  numbered 
the  same  as  in  the  elevation.  With  D"  of  Fig.  504  as 
a  center,  from  each  of  these  points  draw  arcs  indefi- 
nitely to  the  left,  as  shown.  Upon  the  arc  drawn  from 
IF  set  off  spaces  corresponding  to  those  used  in  spac- 
ing the  plan,  beginning  with  IF,  as  shown  by  the  small 
figures,  and  from  each  point  draw  a  line  toward  the 


center  D"  cutting  arcs  of  corresponding  number  drawn 
from  the  line  F"  IF.  A  line  tra'ced  through  the 
points  of  intersection  (g  to  F2)  will  give  the  shape  of 
the  bottom  of  the  cone  to  fit  against  the  side  of 
one  of  the  gables,  or  one-eighth  of  the  complete 
pattern. 

By  repeating  the  space  1  5  upon  the  arc'  drawn 
from  H1  seven  times  additional,  as  marked  by  the  points 
lando,  the  point  V  will  be  reached,  from  which  aline 
drawn  to  D2  will  complete  the  envelope  of  the  cone. 
From  the  points  marked  1  and  5  draw  lines  toward  D1 
intersecting  the  arcs  of  corresponding  number.  This 
will  locate  all  of  the  highest  and  lowest  points  of  the 
pattern,  after  which  the  miter  cut  from  g  to  F'  can  be 
transferred  by  any  convenient  means,  as  shown  from  <j 
to/,  andso  on,  reversing  it  each  time,  as  shown.  In  the 
case  of  a  spire  of  very  tall  and  slender  proportions  it 
will  be  sufficiently  accurate  for  practical  purposes  to 
draw  the  lines  //  F'  and  g  /  straight.  But  the  broader 
the  cone  becomes  at  its  base  the  more  curved  will  tin- 
line  g  FJ  become.  With  a  radius  equal  to  D  E  of  I-'i^. 
503  describe  the  arc  E3  F,  as  shown,  which  will  com- 
plete the  pattern. 


PROBLEM    151. 


The  Pattern  of  a  Conical  Spire  Mitering  Upon  Eight  Gables. 


In  Fig.  505  is  shown  the  elevation  of  a  pinnacle 
having  eight  equal  gables,  upon  which  the  conical 
spire  E  F  -P  I  is  to  be  fitted.  Produce  the  sides  F  E 
and  P  I  until  they  meet  in  the  point  D,  which  is  the 
apex  of  the  spire.  Let  A  H'  S  K  M  N1  T  U  represent 
the  plan  of  the  pinnacle  drawn  in  line  just  below  the 
elevation.  To  ascertain  the  length  of  the  cone  forming 
the  spire  at  its  longest  points,  where  it  terminates  in 
the  valleys  between  the  gables,  it  will  be  necessary  to 
construct  a  section  on  the  line  A  B  representing  one  of 
the  valleys  in  plan,  which  can  be  done  as  follows : 
From  the  points  D,  F  and  H  in  the  elevation  project 
lines  horizontally  to  the  left,  which  intersect  with  any 
vertical  line,  as  D1  B1,  representing  the  center  line  of 
spire  in  the  section.  Upon  the  line  drawn  from  H  set 
off  from  B'  a  distance  equal  to  A  B  of  the  plan  and 


draw  A'  F'.  From  D'  draw  a  line  parallel  to  D  F  cut- 
ting A1  F'  in  R' ;  then.D'  R1  will  be  the  length  or  slant 
hight  of  the  cone  at  its  longest  points,  and  a  line  from 
R1  projected  back  into  the  elevation  will  locate  the  base 
of  the  cone  in  that  view,  as  shown  at  R. 

From  B  as  a  center,  with  a  radius  equal  to  R3  R1, 
describe  a  circle  in  the  plan,  which  will  represent  the 
base  or  plan  of  the  cone.  Divide  an  eighth  of  this 
circle  into  any  number  of  equal  parts,  as  shown  by  1, 
2,  3,  4  and  5,  which  spaces  are  to  be  used  in  measur- 
ing off  the  arc  circumscribing  the  pattern.  Draw  any 
line,  as  D  R  in  Fig.  506,  upon  which  set  off  the  sev- 
eral points  in  the  line  D  R,  as  shown  by  the  letters, 
and  from  D  as  center  describe  arcs  indefinitely  from 
each  point,  as  shown.  On  the  arc  drawn  from  R  stop 
off  spaces  corresponding  to  one-eighth  of  the  plan,  as 


Pattern  Problems. 


283 


shown  by  1,  2,  3,   4  and  5.      Draw  the  line  U  5,   as 
shown,  cutting  the  are  from  F,  as   indicated  byy!    By 


the  arc  R  represents  the  line  of  points  in  the  base  of 
the  cone  to  fit  down  between  the  gables.  Therefore 
from  F  to  the  middle  point  3  draw  F  </,  and  from  f 
draw/ y.  Then  T)  f  y  F  will  be  one-eighth  of  the  re- 
quired pattern.  Set  the  dividers  to  ]  3  on  the  arc  R 
ami  step  of  I  a  sufficient  number  of  additional  spaces  to 
complete  the  pattern,  as  shown  by  3,  5,  3,  5,  etc.,  to  W. 
Draw  W  D.  Also  from  the  points  3  draw  the  lines 
/'.</  and  </  /',  thus  completing  the  pattern. 

In  case  the  work  is  of  large  dimensions  it  will  be 
advisable  to  miter  the  cone  from  F  to  g,  y  to/,  etc., 
in  the  manner  shown  in  the  preceding  problem,  but  in 


Fig.  SOS. — Eleeatiun  and  Plan  of  Conical  Spire  Mitering  upon 
Eirjht  Gables. 


inspection  of  the  elevation  it  will  be  seen  that  the  arc 
F  represents  the  line  of  the  top  of  the  gables,  and  that 


Fig.  506.— Pattern  of  Spire  Shown  in  Fig.  505. 


case  the  work  is  small  it  will  be  sufficiently  accurate  to 
make  the  lines  fy  straight,  as  shown. 


284 


The  New  Metal   Worker  Pattern  Book. 


PROBLEM    152. 


Patterns  for  a  Two-Piece  Elbow  in  a  Tapering:  Pipe. 


In  the  solution  of  this  problem  two  conditions  may 
arise;  in  the  first,  the  two  pieces  of  the  elbow  have  the 
same  flare  or  taper,  while  in  the  second  case  one  of  the 


both  would  present  to  view  two  perfect  ellipses  ot  ex- 
actly the  same  proportions  and  dimensions;  and,  there- 
fore, that  if  the  two  parts  be  placed  together  again, 


I  I     I      I  1 

iii  ii 


Fig.  507.— A  Two-Piece  Elbow  in  a  Tapering  Pipe. 


pieces  may  have  more  flare  than  the  other.  It  has 
been  shown  in  the  chapter  on  geometrical  problems  that 
an  oblique  section  through  the  opposite  sides  of  a  cone 
is  a  perfect  ellipse.  Keeping  this  in  mind,  it  is  evident 
that  if  the  cone  shown  by  ABC  in  PV.  507  were 
made  of  some  solid  material  and  cut  obliquely  by  the. 
plane  D  E  and  the  severed  parts  placed  side  by  side, 


turning  the  upper  piece  half  way  around,  as  shown  by 
D  E  A',  the  edges  of  the  two  pieces  from  D  to  E  would 
exactly  coincide. 

Taking  advantage  of  this  fact,  then,  it  only  becomes 
necessary  to  ascertain  the  angle  of  the  line  D  E,  neces- 
sary to  produce  the  required  angle  between  the  two 
pieces  of  an  elbow,  both  of  which  have  equal  flare. 


Path i' 1 1 


285 


Therefore,  at  any  convenient  point  upon  the  axis  A  11, 
as  I,  draw  I  ,7  at  the  angle  which  the  axis  <>[  the  upper 
piece  is  iv quired  to  m,-ike  with  that  »f  the  lower,  then 
bisect  the  angle  J  1  II.  as  shown  l>v  the  line  I  K. 
I>raw  1)  K  p;.ra!!"|  to  I  \\  at  the  required  hight  of  the 
lower  piece,  which  will  lie  the  miter  line  sought. 

Before  completing  the  elevati >f  1  he1  elbow  it. 

will  lie  necessarv  to  notice  a  peculiarity  of  the  oblique 
section  of  a  eonc — vi/..  that  although  the  line  A  II 
bisects  the  cone  and  its  base  it  does  not  bisect  the 
oblique  line  1)  K,  as  by  measurement  the  center  of  1)  K 
is  found  to  be  at  x.  Therefore,  through  the  point/;, 
which  is  as  far  to  the  right  of  x  as  point  "  is  to  the  left 
of  it,  draw  auv  line.  08  />  A',  parallel  to  I  .7  and  make  l> 
A'  equal  in  length  to  a  A,  and  draw  A1  I>  and  A1  E. 
Xcxt  draw  (!'  K'  at  right  angles  to  A'  />.  representing 
the  upper  end  of  the  elbow.  Make'  1)  F  equal  In  K  K1. 
and  K  (i  equal  to  D  G'.  Then  B  F  (I  0  will  be  the 
elevation  of  a  frustum  of  a  cone,  which,  when  cut  in 
two  upon  the  line  D  K,  will,  when  the  upper  section  is 
turned  half-way  around  upon  the  lower  part,  form  the 
elbow  BDG1  F1  K  ('. 

At  an  v  convenient  distance  below  the  base  of  cone 
B  0  draw  half  the  plan,  as  shown  by  lj  II  M,  which 
divide  into  any  convenient  number  of  equal  spaces. 
From  the  points  of  division  erect  lines  vertical! v,  cut- 
ting the  base  of  the  cone  13  C,  and  thence  carry  them 
toward  the  point  A,  cutting  the  miter  lino  D  E.  Plac- 
ing the  T-square  parallel  to  the  base  line  B  C  bring  it 
successively  against  the  points  in  D  E,  cutting  the  side 
of  the  cone,  as  shown  below  D. 

From  A  as  center,  with  radii  A  15  and  A  F,  draw 
ares,  as  shown.  Upon  the  are  drawn  from  B,  begin- 
ning at  any  convenient  point,  as  N,  step  off  a  stretch- 
out of  L  II  M,  as  shown  by  the  small  figures.  From 
each  of  the  points  thus  obtained  draw  measuring  lines 


toward  the.  point  A,  and  from  the  last  point  0  one 
cutting  tli;-  are  drawn  from  F  at  <^.  Placing  one  point 
of  the  compasses  at  the  point  A,  bring  the  pencil  point 
in  turn  to  each  of  the  points  in  the  side  of  the  cone 
below  D  and  cut  measuring  lines  of  corresponding 
number.  Then  a  line  traced  through  the  points  of  in- 
tersection, as  shown  from  S  to  It,  will  be  the  miter  cut 
between  the  two  parts  of  the  pattern  of  the  frustum  0 
N  I'  Q  necessary  to  form  the  patterns  of  the  required 
elbow. 

As  but  half  the  plan  of  the  cone  was  used  in  ob- 
taining a  stretchout,  the  drawing  shows  but  halves  of 
the  patterns.  In  duplicating  the  halves  to  form  the 
co  uplete  patterns  the  upper  piece  can  be  doubled  upon 
the  line  Q  S  and  the  lower  part  upon  the  line  R  N, 
thus  bringing  the  joints  on  the  short  sides. 

If,  according  to  the  second  condition  stated  at  the 
beginning  of  this  problem,  the  upper  section  of  this 
elbow  is  required  to  have  more  or  less  flare  than  the 
lower  section,  thereby  placing  the  apex  A1  nearer  to  or 
farther  away  from  the  line  D  E,  a  different  course  will 
have  to  be  pursued  in  obtaining  the  pattern.  If,  for 
instance,  the  hight  of  the  cone  A  II  be  reduced,  the 
1  base  BC  remaining  the  same,  the  proportions — that  is, 
the  comparative  length  and  width — of  the  ellipse  derived 
from  the  cut  I)  K  would  be  different  from  thosederived 
from  the  same'  cut  were  the  proportions  of  the  cone  to 
remain  unchanged.  Therefore,  since  the  shape  of  the 
lower  piece  at  the  line  D  E  is  a  fixed  factor,  if  the 
circle  at  G1  F1  be  shifted  up  or  down  the  axis,  or,  re- 
maining where  it  is,  its  diameter  be  changed,  the  piece 
D  G'  F1  E  becomes  an  irregular  tapering  article,  in 
which  case  its  pattern  can  most  easily  be  obtained  by 
triangulation.  Patterns  for  pieces  embodying  those 
conditions  can  be  found  in  Section  3  of  this  chapter, 
to  which  the  reader  is  referred. 


PROBLEM    153. 


Patterns  for  a  Three-Piece  Elbow  in  a  Tapering:  Pipe. 


In  Fig.  508  is  shown  a  three-piece  elbow  occur- 
ring in  taper  pipe,  in  which  the  flare  is  uniform 
throughout  the  three  sections.  In  solving  this  problem 
the  simplest  method  will  be  to  construct  the  elevation 
of  the  elbow  and  an  elevation  of  an  entire  cone,  from 
which  several  sections  may  be  cut  to  form  the  re- 


quired elbow,  at  one  and  the  same  time.  Therefore 
in  the  ele\ation  of  the  cone  E  F  G  let  L1  M'  be  drawn 
at  a  distance  from  E  F  equal  to  the  total  length  of  the 
three  pieces  measured  upon  their  center  lines,  and 
also  let  its  length  be  equal  to  the  diameter  of  the 
elbow  at  its  smaller  end ;  then  through  E  and  L1  and 


286 


Tlte  Xew  Metal    Worker  Pattern  Book. 


through  F  and  M'  draw  the  sides  of  the  cone,  inter- 
secting in  G. 

At  any  convenient  point,  as  B,  draw  the  line  B  A 
at  the  angle  which  the  axis  of  the  middle  piece  is 
required  to  make  with  that  of  the  lower  (in  this  case 
45  degrees),  and  bisect  the  angle  ABC,  as  shown,  by 
the  line  B  D.  Parallel  with  B  D  draw  P  B  at  any  re- 


upper  piece  is  required  to  make  with  that  of  the 
middle  piece  (in  this  case  also  45  degrees,  or  hori- 
zontal), and  bisect  the  angle  S  U  I,  as  shown,  by  U  T. 
Then  the  miter  line  N  0  can  be  drawn  parallel  with 
U  T  at  any  required  distance  from  I,  upon  which 
locate  the  point  i,  making  N  i  equal  to  O  j.  From  i 
draw  the  axis  of  the  upper  piece  of  the  elbow  paral- 


Fiy.  508.— A  Three-Pica  Elbow  in  a  Tapering  Pipe. 


quired  -night,  upon  which  locate  the  point  I  making  P  I 
equal  to  E  J.  (The  reason  for  this  is  explained  in  the 
previous  problem.)  From  the  point  I  draw  the  axis 
of  the  second  section  of  the  elbow  parallel  with  A  B, 
making  it  (I  H)  equal  to  J  G,  and  draw  P  II  and  R  H. 
From  any  convenient  point  upon  this  axis,  as  U, 
draw  U  S  at  the  required  angle  which  the  axis  of  the 


.lei  with  U  S,  making  i  K  equal  to  j  H.  Next  locate 
the  line  N  0  upon  the  original  cone,  making  N'  P 
equal  to  0  R  and  0'  R  equal  to  P  N.  Now  make 
N  L  equal  to  N'  L  and  M  O  equal  to  M'  0'  and  draw 
M  L. 

It  may  be  remarked  -here  that  on  account  of  the 
shifting  of  the  positions  of  the  axes  of    the  several 


Pattern  PrMems, 


287 


pieces  upon  the  miter  lines  l>y  turning  them,  as  shown 
by  L  ,1  and  i  j,  it  will  lie  impossible  to  ascertain  with 
extreme  aeeiiraev  the  lengths  of  the  various  pieces 
upon  their  axes  until  the  elevation  E  P  N  L  M  O  R  F 
is  drawn,  and  therefore'  to  obtain  the  position  which 
the  line  M  L  will  occupy. 

This  method  of  solving  the  problem  is  given  upon 
the  supposition  that  its  simplicity  will  compensate  for 
this  slight  inaccuracy,  as  usually  ditl'erences  of  length 
can  be  made  up  in  the  parts  with  which  the  elbow  may 
be  connected.  If  the  lines  M  h  and  K  F  are  to  be 
assumed  at  the  outset  as  lixed  factors  between  which 
a  tapering  elbow  is  to  be  constructed,  it  will  be  some- 
what difficult  to  ascertain  the  exact  dimensions  of  a 
cone,  K  F  G,  which  can  be  cut  and  its  parts  turned  so 
as  to  constitute  the  required  elbow.  Hence,  while 
two  of  the  pieces  (say  the  two  lower  ones)  can  easily 
be  cut  from  an  entire  cone  assumed  at  the  outset,  the 
third  piece  will  have  to  be  drawn  arbitrarily  to  fit  be- 
tween the  last  miter  line  N  O  and  the  small  end  M  L, 
and  will  very  likely  be  of  different  flare  from  that  of 
the  other  two  pieces.  This  will  necessitate  the  last 
section  being  cut  by  the  method  of  triangulation, 
problems  in  which  are  demonstrated  in  Section  3  of 
this  chapter,  to  which  the  reader  is  referred. 


Having,  as  explained  above,  obtained  the  lines 
of  cut  through  the  cone,  the  patterns  may  be  described 
as  follows :  Draw  the  plan  V  W  Y,  its  center  X  fall- 
ing upon  the  axis  of  the  cone  produced,  which  divide 
in  the  usual  manner  into  any  convenient  number  of 
equal  parts.  Through  the  points  thus  obtained  erect 
perpendiculars  to  the  base  E  F,  and  thence  carry  lines 
toward  the  apex  G,  cutting  the  miter  lines  P  R  and 
N'  O'.  With  the  T-square  at  right  angles  to  the  axis 
G.C,  and  brought  successively  against  the  points  in 
N'  U'  and  P  R.  cut  the  side  G  F  of  the  cone,  as  shown 
by  the  points  above  0'  and  below  R.  From  Gas  center, 
with  radius  G  F,  describe  the  arc  E1  F1,  upon  which  lay 
off  the  stretchout  of  the  plan  V  W  Y,  as  shown  by  the 
small  figures  1,  2,  3,  etc.,  and  from  these  points  draw 
measuring  lines  to  the  center  G.  From  G  as  center 
describe  arcs  corresponding  to  the  distance  from  G  to 
the  several  points  established  in  G  F,  which  produce 
until  they  intersect  lines  of  corresponding  numbers 
drawn  from  the  center  G  to  the  arc  E1  F1.  Through 
these  points  of  intersection  trace  lines,  as  shown  by 
0''  N2  and  P'  R1.  From  G  as  center,  with  radius  G  M', 
describe  the  arc  L'  W.  Then  L'  M'  N3  0'  is  the  half 
pattern  of  the  upper  section,  0'  N3  R'  P'  that  of  the 
middle  section,  and  P'  R'  F'  E'  that  of  the  lower  section. 


PROBLEM  154. 


The  Patterns  for  a  Regular  Tapering  Elbow  in  Five  Pieces. 


Fig.  509.— Diagram  of  Angles  for  a  Five  Piece  Elbow. 


In  this  problem,  as  in  the  two  immediately  preced- 
ing, the  various  pieces  necessary  to  form  the  elbow 
may  be  cut  from  one  cone,  whose  dimensions  must  be 
determined  from  the  dimensions  of  required  elbow. 
The  first  essential  will  be  to  determine  the  angle  of  the 
cutting  lines,  which  may  be  done  the  same  as  if  the 
elbow  were  of  the  same  diameter  throughout. 

Such  an  elbow  of  five  pieces  would  consist  of  three 
whole  pieces  and  two  halves;  therefore,  if  it  is  to  be  a 
right  angle  elbow,  divide  any  right  angle,  as  A'  B  C'  in 
Fig.  509,  into  four  equal  parts,  as  shown  by  the  points 
1,  '2,  3.  Bisect  the  part  A'  B  3  by  the  line  A  B  and 
transfer  the  portion  A'  B  A  to  the  opposite  side  of  the 
figure,  as  shown  by  C'  B  C. 

This  gives  the  right  angle  ABC  divided  into  the 
same  number  of  pieces  and  half -pieces  as  would  be  em- 
ployed in  constructing  an  ordinary  five-piece  elbow, 


288 


'fie  Xcw  Metal    Worker  l*uUern  Book. 


M 


Fig.  BIO.— A.  Fivt-Pieei  Elbow  in  a  Taptriny  fife. 


Pattern  Problems. 


The  division  lines  in  this  diagram  arc  of  the  correct 
alible  for  the  miter  lines  in  the  el  how  pattern,  and  there- 
fore can  be  used  upon  the  diagram  of  the  cone,  out  of 
which  are  to  be  obtained  the  pieces  to  compose  the  re- 
quired elbow. 

It  is  assumed  that  the  amount  of  rise  and  projec- 
tion are  not  specified,  therefore  after  having  got  the 
line  of  the  angle  or  miter  it  becomes  a  matter  of  judg- 
ment upon  the  part  of  the  pattern  cutter  what  length 


Fig.  all.— Elevation  of  Five-Piece  Tapering  Elbow. 

shall  be  given  to  each  of  the  pieces  composing    the 
elbow. 

In  Fig.  510,  let  A  B  represent  the  diameter  of  the 
large  end  of  the  elbow.  From  the  middle  point  in  the 
line  A  B,  as  C,  erect  a  perpendicular  line,  as  indicated 
by  C  N,  producing  it  indefinitely.  On  the  line  C  N, 
proceeding  upon  judgment,  as  already  mentioned,  set 
off  C  X  to  represent  the  length  of  the  first  section  of 
the  elbow  measured  upon  its  center  line.  With  X  thus 
determined,  draw  through  it  the  line  D  E,  giving  it  the 
same  angle  with  A  B  as  exists  between  B  C1  of  Fig. 
509  and  the  horizontal  B  C.  This,  in  all  probability, 
can  most  readily  be  done  by  extending  B  A  indefinitely 


beyond  A  and  letting  E  D  intersect  with  B  A  extended, 
producing  at  their  intersection  an  angle  equivalent  to 
C  B  C1  of  Fig.  509.  From  the  point  X  set  off  the  dis- 
tance X  Y,  also  established  by  judgment,  thus  deter- 
mining the  position  across  the  cone  of  the  rniterlineof 
the  next  section.  Through  Y  draw  G  F  at  the  sam< 
angle  as  D  E,  already  drawn,  but  inclined  in  the  oppo 
site  direction.  In  like  manner  locate  the  two  other 
miter  lines  shown  in  the  diagram,  finally  obtaining  the 
point  Z.  From  Z  set  off  the  width  toward  N  of  the 
last  section  of  the  pattern,  and  through  the  point  N" 
tli us  obtained  draw  the  line  M  0  at  right  angles  to  ON, 
making  it  in  length  equal  to  the  diameter  of  the  small 
end  of  the  elbow  and  placing  its'  central  point  at  N. 
Through  the  points  A  M  and  B  0  of  the  figure  thus 
constructed  draw  lines,  which  produce  indefinitely  until 
they  intersect  the  axis  in  the  point  P.  Then  P  will  be 
the  apex  of  the  required  cone. 

Construct  a  plan  of  the  base  of  the  cone  or  large 
end  of  the  elbow  below  and  in  line  with  the  diagram, 
as  shown  in  the  drawing,  which  divide  into  any  con- 
venient number  of  spaces,  as  indicated  by  the  small 
figures,  and  from  the  points  thus  obtained  carry  lines 
vertically,  cutting  the  base  of  the  cone  A  B.  From 
A  B  continue  them  toward  the  apex  of  the  cone,  cut- 
ting the  several  miter  lines  drawn.  With  the  apex  P 
of  the  cone  for  center,  and  with  P  B  as  radius,  describe 
the  arc  T  U,  upon  which  eet  off  a  stretchout  of  one- 
half  the  plan,  all  as  indicated  by  the  small  figures. 
From  the  points  thus  established  in  T  U  carry  lines  to 
the  center  P.  With  the  T-square  placed  at  right  angles 
to  the  axis  N  C  of  the  cone,  and  brought  against  the 
points  of  intersection  in  the  several  miter  lines  made 
by  the  lines  drawn  from  points  in  the  base  of  the  cone 
to  the  apex,  cut  the  side  O  B  of  the  cone,  as  shown. 
Then  from  P  as  center,  with  radii  corresponding  to  the 
distance  from  P  to  the  several  points  on  0  B,  as  men- 
tioned, strike  arcs  cutting  the  lines  of  corresponding 
numbers  in  the  pattern  diagram,  as  shown.  Then  lines 
traced  through  the  points  thus  obtained,  as  indicated  bv 
D'  E',  F1  G',  etc.,  will  cut  the  pattern  0  W  U  T  of  the 
frustum  in  such  a  manner  that  the  sections  will  consti- 
tute the  half  patterns  of  the  pieces  necessary  to  form 
the  required  elbow.  In  Fig.  511  is  shown  an  elevation 
of  the  elbow  resulting  from  the  preceding  operation. 


Til,'      Xl'lf      .]/,'/,(/       II'W.VV      ['llllr 


PROBLEM    155. 

The  Frustum  of  a  Cone  Intersecting  a  Cylinder  of  Greater  Diameter  than  Itself  at  Other  than 

Right  Angles. 


la  Fig.  512,  K  (i  II  F  iv prose n Is  an   elevation   of    I   obtained  drop  points  parallel  with  the  side  (!   II  <>[   the 


the  cylinder,  and  M  N  L  K  an  elevation  of  the  frustum 
of  a  cone  intersecting  it.  V  Z  Q  represents  the  profile 
or  plan  of  the  cylinder,  to  which  it  will  l>e  necessary  to 
add  a  correctly  drawn  plan  of  the  frustum  before  the 
miter  lino  in  elevation  can  be  obtained.  At  any  con- 


cylinder  and  continue  them  indefinitely,  cutting  the 
line  F1  ()',  which  is  drawn  through  the  center  of  the 
plan  of  the  cylinder  at  right  angles  to  the  elevation,  all 


as  shown  in  the  ciurravinir.       Make  V  \V 


V  \\" 


of  the  lirst  section  c  .instructed,     hi  like  manner  meas 


C    E 


uis    12     n    10     a      'us     5^      12    x 


xf-—i-^f- 

iF 


W' 


Fig.  512.— The  frustum  of  a  C/'une  Iiitcrseetiiiy  a,  Cylinder  uf  Itiviter  Diameter  than  Itself  at.  Other  than  Itirjkt 


venient  point  on  the  axial  line  T  O  of  the  cone  construct 
the  profile  V  Y  X  W,  which  represents  a  section 
through  the  cone  on  the  line  M  N.  •  Divide  the  section 
V  Y  X  W  into  any  convenient  number  of  equal  spaces 
in  the  usual  manner,  as  shown  by  the  small  tignres  1. 
2,  3,  4,  etc.  From  each  of  the  points  thus  established 
drop  lines  parallel  with  the  axis  of  the  cone  cutting 
the  line  M  N,  From  the  intersections  in  M  N  thus 


distances  from  the  center  line  V  X  of  the  iirst  section 
to  the  points  2,  3,  4-,  etc.,  and  set  oft  corresponding 
spaces  in  the  plan  view,  measuring  from  M' N',  upon 
lines  of  corresponding  numbers  dropped  from  the  in- 
tersections in  M  N,  already  described.  Then  a  line 
traced  through  these  points  will  represent  a  view  of  the 
upper  end  of  the  frustum  as  it  would  appear  when 
looked  at  from  a  point  directly  above  it.  Produce  the 


21U 


sides  <>!'  the  frustiua  K.  .\[  ;uid  L  N  until  they  meet  in 
tlie  point  <  I.  From  (.)  drop  ;i  line  parallel  to  the  side 
(r  II  of  the  cylindt.'r,  cutting  the  line  K1  ()'  in  the  point 

0',  thus  establishing  the  positi< f  the  apex  of  the 

cone  in  the  plan.  From  the  point  O'  thus  estaMislied 
draw  lines  through  the  several  points  in  the  section 
M'  V  N'  W.  which  produce  until  they  intersect  the 
plan  of  the  cylinder  in  points  between  Z  and  (.).  as 
shown  in  the  engraving.  From  0,  the  apex  of  the  cone 
in  the  elevation,  draw  lines?  through  the  several  points 
in  M  X  alreadv  determined,  which  produce  until  they 
cross  G  II,  the  side  of  the  cylinder,  and  continue  them 
inward  indefinitely.  Intersect  these  linos  l>v  lines  drawn 
verticallv  from  the  points  of  corresponding  number  be- 
tween Z  and  Q  of  the  plan  just  determined.  Then  a 
line  traced  through  these  intersections,  as  indicated  by 
K  T  L,  will  represent  the  miter  between  the  frustum 
and  cylinder,  as  seen  in  elevation. 

To  lav  off  the  pattern  proceed  as  follows:  From 
O  as  center,  with  O  N  as  radius,  describe  the  are  P  R, 
on  which  set  oft  a  stretchout  of  the  section  Y  V  W  X 
in  the  usual  manner.  From  0,  through  the  several 
points  in  P  It  thus  obtained,  draw  radial  lines  indefi- 
nitely. From  the  several  points  in  the  miter  line  K  T  L 


draw  lines  at  right  angles  to  the  axis  O  T  of  the  cone. 
producing  them  until  they  cut  the  sick:  N  L.  From  Oas 
center,  with  radii  corresponding  to  the  distance  from  0  to 
the  several  points  in  N  L  just  obtained,  describe  ares, 
which  produca until  they  intersect  radial  I in/s  of  corre- 
sponding number  drawn  through  the  stretchout  P  R. 
Then  a  line  traced  through  these  points  of  intersection,  as 
indicated  by  S  17  U,  will  be  the  lower  line  of  the  pat- 
tern sought,  and  P  S  L'  II  R  will  be  the  complete  pattern. 
The  pattern  for  the  cylinder  and  the  opening  in 
the  same  to  lit  the  intersection  of  the  cone  is  reallv  a 
problem  in  parallel  forms,  with  which  problems  (Section 
1)  it  should  properly  be  classed.  F'  Z  Q  is  the  pro- 
file of  the  cylinder,  and  L  T  K  is  the  miter  line.  The 
stretchout  R  1)  is  drawn  at  right  angles  to  K  F,  the 
direction  of  the  mold  or  cylinder.  The  points  between 
Z'  and  Q"  of  the  stretchout  are  duplicates  of  those  be- 
tween Z  and  Q  of  the  plan.  Place  the  f-square  at 
right,  angles  to  the  cylinder,  and,  bringing  it  successively 
against  the  points  in  the  miter  line  K  T  L,  cut  lines  of 
corresponding  numbers.  A  line  traced  through  the 
points  of  intersection  thus  formed,  as  shown  by  Z'  K' 
Q'  L',  will  be  the  shape  of  the  required  opening  in  the 
cylinder. 


PROBLEM  156. 

The  Patterns  of  the  Frustum  of  a  Cone  Joining:  a  Cylinder  of  Greater  Diameter  than  Itself  at  Other 
than  Right  Angles,  the  Axis  of  the  Frustum  Passing  to  One  Side  of  That  of  the  Cylinder, 


Let  E  F  II  G  in  Fig.  513  be  the  elevation  of  a 
cylinder,  which  is  to  be  intersected  by  a  cone  or  frus- 
tum, 1)  A  .1  C,  at  the  angle  F  D  A  in  elevation,  and 
which  is  to  be  set  to  one  side  of  the  center,  all  as 
shown  by  S  PL  M  R  of  the  plan.  Opposite  the  end 
of  the  frustum,  in- both  elevation  and  plan,  draw  a  sec- 
tion of 'it,  as  shown  by  T  I'  V  W  in  the  elevation  and 
T1  I*'  V  W  in  the  plan.  Divide  both  of  these  sec- 
tions into  the  same  number  of  equal  parts,  commenc- 
ing at  corresponding  points  in  each,  and  number  them 
as  shown  by  the  small  figures  in  the  diagram.  From 
the  points  in  T  I"  V  \V  carry  lines  parallel  to  the  axis 
of  the  cone,  cutting  the  line  A  .1,  and  thence  drop 
them  vertically  across  the  plan.  From  the  points  in 
the  section  T1  U'  V  \V  draw  lines  parallel  to  the  axis 
of  the  cone,  as  seen  in  plan,  intersecting  the  lines  of  cor- 
responding number  dropped  from  A  J  just  described. 
Through  these  points  of  intersection1  trace  a  line,  as 
shown  by  L  M.  Then  L  M  wdl  show  tin  end  of  the 


frustum  A  J  as  it  appears  in  plan.  From  X,  the 
apex  of  the  cone  in  elevation,  drop  a  line  vertically, 
cutting  the  axis  of  the  cone  in  plan  as  shown  at  X'. 
From  X  draw  lines  through  the  points  in  A  J  and 
extend  them  through  the  side  of  the  cylinder  indefi- 
nitely. From  X'  through  the  points  in  L  M  draw 
lines  cutting  the  plan  of  the  cylinder,  as  shown  from 
P  to  R,  and  from  these  points  carry  lines  vertically, 
intersecting  those  of  corresponding  number  in  the 
elevation  drawn  from  the  apex  X.  Then  a  line  traced 
through  these  points,  as  shown  by  D  K  C  N,  will  be 
the  miter  line  in  elevation. 

For  the  pattern  of  the  frustum,  from  X  as  center, 
wi.th  radius  X  A,  describe  the  arc  A1  J',  upon  which 
lay  off  a  stretchout  of  the  section  T  U  V  W,  through 
the  points  in  which,  from  X,  draw  radial  lines  indefi- 
nitely. From  the  points  in  1)  K  C  N  carry  lines  at 
right  angles  to  the  axis  of  the  cone,  cutting  the  side 
A  D  extended,  as  shown  from  D  to  B.  From  X  as 


202 


The  New  Metal  Worker  Pattern  Book. 


center,  with  radii  corresponding  to  the  distance  from 
X  to  the  various  points  in  the  line  D  B,  describe  arcs 
cutting  radial  lines  of  corresponding  number  in  tho 
pattern.  Through  the  points  of  intersection  in  the 


tern  of  the  frustum  D  A  J  C,  inhering  with  the  cylin- 
der at  the  angle  described. 

The  method  of  obtaining  the  pattern  of  the  cyl- 
inder is  analogous  to  that  described  in  the  preceding 


Fig.  SIS. — The  Patterns  of  the  Frustum  of  a  Cone  Joining  a  Cylinder  of  Greater  Diameter  than  Itself  at  Other  than  Right  Angles 
the  Axis  of  the  Frustum  Passing  to   One  Side   of   that  of  the  Cylinder. 

pattern    thus    obtained    trace    a    line,    as     shown    by      problem,    and    is    clearly    shown    at    the    left    in    the 


C1  K1  N1  D1.      Then  D'  N1  K1  C'  A1  J1  will  be  the  pat- 


drawing. 


PROBLEM   157. 

The  Patterns  of  a  Cone  Intersected  by  a  Cylinder  of  Less  Diameter  than  Itself,  Their  Axes  Crossing  at 

Right  Angles. 


In  Fig.  514,  let  B  G  E  D  F  A  C  be  the  elevation 
of  the  required  article.  Draw  the  plan  in  line  with  the 
elevation,  making  like  points  correspond  in  the 


views,  as  shown  by  M  0  S  T  U  P  N".  Let  D  M  E  be  a 
half  Section  of  the  cylinder  in  the  elevation  and  D1  M1 
E1  a  corresponding  section  in  the  plan. 


Pattern  Problems. 


293 


Divide  these  sections  into  any  convenient  number 
of  equal  parts,  commencing  at  the  same  point  in  each, 
as  shown  by  the  small  figures,  and  draw  the  center  line, 
of  the  cylinder  in  plan  D'R.  From  each  of  the  points  in 
the  section  shown  in  elevation  carry  lines  parallel  to  D 
H 


Fig.  S14.—A  Cone  Intersected  by  a  Cylinder  of  Less  Diameter  than 
Itself  at  Right  Angles  to  Its  Axis. 

F  cutting  the  side  of  the  cone,  and  extend  them  some 
distance  jnto  the  figure  for  further  use.  From  the 
several  points  of  intersection  with  the  side  of  the  cone, 
as  shown  by  a,  b,  c,  d  and  e,  drop  lines  parallel  to  the  axis 


of  the  cone  cutting  the  line  D'  R  of  the  plan,  giving  the 
points  a,b',c\dl  and  el,  and  through  each  of  these  points, 
from  R  as  center,  describe  an  arc,  as  indicated  in  the  en- 
graving. From  the  points  in  the  profile  D'  M1  E1  of 
the  plan  draw  lines  parallel  to  the  sides  of  the  cylinder, 
producing  them  until  they  meet  the  arcs  drawn  through 
corresponding  points,  giving  the  points  indicated  by 
I1,  2',  3',  4'  and  5'.  From  these  points  carry  lines 
vertically  to  the  'elevation,  producing  them  until  they 
meet  the  lines  drawn  from  points  of  corresponding 
numbers  in  the  profile  of  the  cylinder  in  the  elevation, 
giving  the  points  T,  2%  3',  4a  and  5'.  A  line  traced 
through  these  points,  as  shown  from  G  to  F,  will  be 
the  miter  line  in  elevation  formed  by  the  junction  of 
the  cvlinder  and  the  cone. 


Fig.  515.— Half  Pattern  of  the  Cone  Shown  in  Fig.  514. 

To  obtain  the  envelope  of  the  cone  with  the  open- 
ing to  fit  the  intersecting  cylinder  proceed  as  follows: 
From  any  convenient  point,  as  A1,  Fig.  515,  draw  A1 
B1,  in  length  equal  to  A  B  of  the  elevation.  Set  off 
points  e',  d',  c",  b'  and  a'  in  it,  corresponding  to  e,  d,  c, 
b  and  a  of  A  B,  Fig.  514.  From  A1  as  center,  with 
radius  A'  B1,  describe  the  arc  B'  V,  upon  which  lay  off 
the  stretchout  of  the  plan  of  the  cone,  as  indicated  by 
the  small  figures  outside  of  the  pattern.  (But  one-half 
of  the  envelope  of  the  cone  is  shown  in  the  engraving.) 
From  the  same  center- A1  describe  arcs  from  the  points 
e",  cT,  c",  V  and  a".  From  the  center  R  of  the  plan  draw 
lines  to  the  circumference  through  the  points  2',  3',  4', 
etc.,  giving  the  points  in  the  circumference  marked  2s, 
33,  43,  etc.  Set  off  by  measurement  corresponding 
points  in  the  arc  B'  V,  as  shown  by  3*,  2',  4*,  5',  etc. 
From  these  points  draw  lines  to  the  center  A1,  inter- 
secting the  arcs  of  corresponding  number  drawn  from 


294 


flic  Ncio  Metal    \Vnrlvr   Pattern 


«%  b\  e",  etc.  A  line  traced  through  these  points  of 
intersection,  as  shown  l>y  K1  U'  G'  P',  will  be  the  shape 
of  the  opening  to  be  cut  in  the  side  of  the  cone  to 
fit  the  mitered  end  of  the  cylinder. 


The  pattern  for  the  cylindrical  part  is  shown  abov 
llii1  elevation,  and   is  obtained    in   accordance  with  111. 
principles  demonstrated  in  the  first  section  of  this  chap- 
ter, which  need  not  be  here  repeated. 


PROBLEM    158. 

The  Patterns  of  a  Cone  Intersected  by  a  Cylinder  of  Less  Diameter  than  Itself  at  Rig-ht  Angles  to  its 
Base,  the  Axis  of  the  Cylinder  Being:  to  one  Side  of  that  of  the  Cone. 


PATTERN  OF  CYLINDER 

In  Fig.  51fi,  let  B  A  C  represent  the  elevation  of 
the  cone,  D  E  Gr  H  the  elevation  of  the  cylinder,  which 
joins  the  cone  at  right  angles  to  the  base  C  C.  J  K 
L  M  N  0  P  Q  is  the  plan  of  the  articles,  which  is  to 
be  drawn  in  line  and  under  the  elevation,  making  like 
points  correspond  in  the  two  views,  as  shown.  Draw 
a  section  of  the  cylinder  in  line  with  the  elevation,  as 
shown  by  E  F  (>  U.  Divide  the  section  of  the  cylinder 
into  any  convenient  number  of  equal  parts,  as  shown 
by  the  small  figures.  From  the  apex  A  drop  a  line 
through  the  plan,  as  shown  by  A  M.  Through  the 
center  of  the  section  of  the  pipe,  as  shown  in  plan, 
draw  a  straight  line  to  the  center  of  plan  of  cone,  as 
shown  by  J  P.  This  line  will  also  be  at  right  angles 
to  K  M.  From  each  of  the  points  in  the  section  of  the 
pipe  in  elevation  drop  lines  parallel  to  the  sides  of  the 
pipe  cutting  the  side  of  the  cone,  extending  them  to 
the  line  J  P  in  plan,  as  shown  by  N  a  ft  c,  etc.  Through 
each  of  these  points,  from  P  as  center,  describe  circles, 


PLAN 

Fill.  Slff.—Plfin  and  Elevation  of  Cnne  Tntersecteil  l>i/  a  Cylinder  at 
RiijM  Anijles  to  its  Base. 


205 


MS  shown,  cutting  the  sides  of  the  plan  of  cylinder. 
From  each  of  the  points  of  intersection  with  the  side 
n|'  the  cone  (A  I!)  draw  lines  parallel  with  the  base,  and 
extend  them  inward.  Lf  it  is  desired  to  show  the  miter 
line  in  elevation  formed  l>v  the  junction  of  pipe  and 
cone,  from  the  points  <l  ef  in  the  plan  of  cylinder  carry 
lines  vertically  to  the  elevation,  producing  them  until 
they  meet  the  horizontal  lines  having  similar  letters 
drawn  through  the  side  of  the  cone  A  B,  giving  the 
points  <j  h,  j.  A  line  traced  through  these1  points,  as 
shown  by  1)  <j  li  j  II,  will  lie  the  miter  line. 


II  f  e  d  I)  <>f  elevation,  as  shown  by  II  r  1>  a  D.  From 
the  center  A  of  pattern  describe  arcs  cutting  the  points 
II  cba  D. 

It.  is  only  necessary  now  to  make  each  of  these  arcs 
ecpial  in  length  to  the  one  to  which'  it  corresponds  in  the 
plan  by  any  method  most  convenient.  Thus  make  ad 
and  n  '/'  equal  to  a  d  of  the  plan,  I  e  and  b  e'  equal  to 
/>  e  of  the  plan  and  c/and  cf  equal  to  c/of  the  plan. 
A  line  traced  through  these  points,  as  shown  by  H  Q 
D  O,  will  be  the  shape  of  the  opening.  Another  method 
of  making  the  measurements  of  the  arcs  is  shown  by 


FitJ.  517. — Half  Pattern  of  Cone  Uluiuiti  in  Fig.  SIS. 


fig.  !il8.  —Perspective  View  of  Cone  and. 
Cylinder  Shown  in  Fig.  510. 


The  half  pattern  of  the  cone,  with  the  opening  to 
tit  the  cylinder,  is  shown  in  Fig.  517,  to  describe  which 
proceed  as  follows:  From  any  convenient  point,  as  A 
in  Fig.  517,  with  A  B  of  Fig.  516  as  radius,  strike  an 
arc  indefinitely,  as  shown.  From  B  of  pattern  set  off 
each  way  the  stretchout  of  .1  M  and  .]  K  of  plan  and 
connect  K  and  M  with  A.  Then  K  A  M  B  is  the  half 
pattern  of  the  cone,  or  as  much  as  shown  on  plan  by 
Iv  .1  M.  To  obtain  the  shape  of  opening  to  be  cut  in 
cone  to  correspond  with  the  shape  of  pipe,  on  A  B,  the 
center  line  of  pattern,  set  off  points  corresponding  to 


the  radial  dotted  lines  of  the  plan  and  pattern  in  the  man- 
ner explained  in  the  problem  immediately  preceding. 

The  pattern  for  the  cylinder  is  obtained  in  the 
manner  usual  with  all  parallel  forms,  its  only  pecul- 
iarity in  this  case  being  that  its  stretchout  is  taken 
from  the  irregular  spaces  upon  the  profile  N  O  P  Q  of 
the  plan,  which  are  transferred  to  the  line  P'  P",  as 

shown. 

A  pictorial  representation  of  the  finished  article  is 
shown  in  Fig.  51 S.  upon  which  some  of  the  lines  of 
measurement  shown  in  Fig.  510  have  been  traced. 


Tlie  New  Metal   Worker  Pattern  BooJc. 
PROBLEM    159. 

Patterns  of  a  Cylinder  Joining:  a  Cone  of  Greater  Diameter  than  Itself  at  Right  Angles  to  the  Side 

of  the  Cone. 


Let  B  A  K  in  Fig.  519  be  the  elevation  of  a  right 
cone,  perpendicular  to  the  side  of  which  a  cylinder,  L 
S  T  M,  is  to  be  joined.  The  first  operation  will  be  to 
describe  the  miter  line  as  it  would  appear  in  elevation. 
Draw  the  section  U  V  of  the  cylinder,  which  divide 
into  any  convenient  number  of  equal  parts,  as  indicated 
by  the  small  figures,  and  from  these  points  drop  lines 
parallel  to  L  S,  cutting  the  side  A  K  of  the  cone  in  the 
points  H,  F  and  D,  producing  them  until  they  cut  the 
axis  A  X  in  the  points  G,  E  and  C.  In  order  to  ascer- 
tain at  what  point  each  of  these  lines  will  cut  the  en- 
velope of  the  cone  it  will  be  necessary  to  construct 
sections  of  the  cone  as  it  would  appear  if  cut  on  the 
lines  G  H,  E  F  and  C  D.  Draw  a  second  elevation  of 
the  cone,  as  shown  by  B'  A1  K',  representing  the  cone 
turned  quarter  way  round ;  the  first  may  be  regarded 
as  a  side  elevation  and  this  as  an  end  elevation.  Draw  a 
plan  under  the  side  elevation  of  the  cone,  as  shown  by 
N  R  P  0,  which  divide  into  any  convenient  number  of 
equal  parts,  and  in  like  manner  draw  a  corresponding 
plan  or  half  plan  under  the  end  elevation,  as  shown  by 
R1  P1  0'.  Divide  this  second  plan  into  the  same  spaces, 
numbering  them  to  correspond  with  the  other  plan. 
From  the  points  1  to  5  in  plan  N  R  P  O  carry  lines 
vertically  to  the  base  B  K  and  thence  toward  the  ;I]M'\ 
A,  cutting  the  lines  C  D,  E  F  and  G  II.  In  like  man- 
ner, from  the  same  points  (1  to  5  inclusive)  in  the  plan 
R'  P1  O1  carry  vertical  lines  to  the  base  B'  K1  and 
thence  toward  the  apex  A'.  Place  the  J-square  at  right, 
angles  to  the  axes  of  the  two  cones,  and,  bringing  it 
against  the  points  of  intersection  of  the  lines  from  X  to 
K  with  C  D,  cut  corresponding  lines  in  the  second  ele- 
vation, and  through  the  points  of  intersection  thus  es- 
tablished trace  a  line,  as  shown  by  C3  C4.  Produce  the 
axis  X1  A1  to  any  convenient  distance,  upon  whicli  set 
off  C1  D1,  in  length  equal  to  C  D,  in  which  set  off  the 
points  corresponding  to  the  points  in  C  D,  and  through 
these  points  draw  lines  at  right  angles  to  C'  D'.  Place 
the  T-square  parallel  to  thev  axis  X1  A',  and,  bringing 
it  against  the  several  points  in  C3  C4,  cut  the  lines  of 
corresponding  number  drawn  through  C'  D',  as  shown, 
and  through  the  intersections  thus  established  trace  a 
line,  as  shown.  Then  C'  D1  is  a  section  of  the  cone  as 
it  would  appear  if  cut  on  the  line  C  D. 

In  like  manner  carry  lines  from  the  points  upon 


E  F  across  to  the  end  elevation,  intersecting  them  with 
lines  of  corresponding  number,  as  shown  from  E3  to  E4, 
and  thence  carry  them  parallel  to  the  axis,  cutting  lines 
drawn  through  E1  F1,  which  with  its  points  has  been 


/•'it/.  '119. — .1  Cylinder  Joininy  a  Cone  of  Greater  Diameter  than  Itxi-lf 
at  Right  Angles  to  the  Side  of  the  Cone. 

made  equal  to  E  F.  The  resulting  profile  E1  F1  is  a 
section  of  the  cone  as  it  would  appear  if  cut  on  the  line 
K  F.  Also  use  the  points  in  <i  II  in  like  manner,  es- 


Pattern  Problems. 


29? 


tablishing  the  profile  G1  fl',  which  represents  a  section 
of  the  cone  as  it  would  appear  if  cut  on  the  line  G  H. 
(Some  of  the  lines  indicating  the  operation  in  connec- 
tion with  sections  E'  F1  and  G'  IT  are  omitted  in  the 
engraving  to  avoid  confusion.) 

[laving  thus  obtained  sections  of  the  cone  corre- 
sponding to  the  several  lines  C  1),  K  F,  G  II,  it  will 
next  be  necessary  to  project  a  plan  of  the  cone,  with 
its  intersecting  cylinder,  at  right  angles  to  L  S,  or  as 
viewed  in  the  direction  of  A  K,  which  plan  shall  in- 
clude all  of  these  sections.  To  do  this  extend  the  line 


Fi(J.  520. — Envelope  of  Cone  Shown  in  Firj.  519. 


A  K  to  a  convenient  distance  above  the  elevation,  and 
project  lines  from  all  other  important  points  parallel  to 
the  same,  as  shown.  At  right  angles  to  A  K  draw  any 
line,  as  C"  V,  as  a  center  line  of  the  new  plan.  As  the 
points  D',  F1  and  H'  of  the  oblique  sections  of  the  cone 
are  all  in  the  line  A  K,  transfer  these  sections  to  the 
new  plan,  so  placing  them  that  their  center  lines  shall 
coincide  with  the  center  line  of  the  new  plan,  and  the 
points  D1,  F'  and  H1  shall  be  at  the  intersection  of  A 
K  with  the  center  line  of  the  plan,  all  as  shown.  Op- 
posite the  end  of  the  cylinder  draw  a  section,  as  indi- 
cated by  U1  V,  which  divide  into  the  same  number  of 
equal  parts  as  used  in  the  divisions  of  U  V,  commenc- 
ing the  division  at  corresponding  points  in  each.  As 
both  halves  of  the  cylinder  and  of  the  cone,  when  di- 
vided by  a  vertical  plane  passing  through  the  axis  of 
each,  are  the  same,  only  one-half  of  the  section  of  the 
cylinder  has  been  numbered.  From  the  points  in  U1 
V  drop  lines  parallel  to  C3  V,  each  line  cutting  its 


corresponding  section,  as  shown  from  x  toy,  and  then 
carry  them  parallel  to  A  K  back  to  the  elevation,  cut- 
ting lines  of  corresponding  number  in  that  view.  That 
is,  from  the  intersection  of  the  line  drawn  from  point 
4  in  U1  V  with  the  profile  C'  L'  cut  the  line  C  D, 
which  in  the  elevation  corresponds  to  the  point  4  in 
the  profile  U  V,  and  from  the  intersection  of  a  line 
drawn  from  3  with  E"  L"  cut  the  line  E  F,  and  so  on, 
all  as  indicated  by  the  dotted  lines.  Then  aline  traced 
through  these  points  of  intersection,  as  shown  by  L  M, 
will  be  the  miter  line  in  elevation,  from  which  the  pat- 
terns may  readily  be  obtained. 

The  intersections  in  the  plan  above  give  all  that  is 
necessary  to  obtain  the  pattern  of  the  cylinder,  which 
can  be  done  as  follows :  Lay  off  a  stretchout  of  the 
profile  U1  V  opposite  the  end  S5  T",  through  the  points 
in  which  draw  the  usual  measuring  lines.  Place 
the  T-square  at  right  angles  to  the  same,  and,  bringing 
it  against  the  points  in  the  miter  line  LM  (or  the  points 
of  intersection  in  x  y  in  the  plan  from  w.hich  L  M  was 
obtained),  cut  the  corresponding  measuring  lines.  Then 
a  line  traced  through  these  points,  as  shown  from  L1  to 
M1,  will  be  the  shape  of  the  pattern  of  the  cylinder  to 
fit  against  the  cone. 

For  the  pattern  of  the  cone  proceed  as  follows : 
From  each  of  the  points  in  the  miter  line  L  M  carry 
lines  horizontally  across,  cutting  the  side  A  B  of 
the  cone,  by  means.of  which  their  distance  from  the 
apex  A  may  be  accurately  measured;  also  through 
these  points  draw  lines  from  the  apex  A  cutting  the 
base  B  K,  continuing  them  vertically  into  the  plan  N 
R  P  O,  as  shown.  It  may  be  noted  that  the  line  from 
point  4  on  L  M  falls  at  point  2  in  the  plan  of  the  cone ; 
likewise  that  the  line  from  3  on  L  M  falls  at  3  in  the 
plan  of  the  cone,  while  the  line  from  2  falls  upon  the 
plan  of  the  cone  at  a  point  marked  a.  From  any 
convenient  point,  as  A2  of  Fig.  520,  with  a  radius  equal 
to  A  B,  describe  the  arc  B2  KJ  B",  which  in  length 
make  equal  to  the  circumference  of  the  plan  of  the 
cone,  setting  off  in  the  same  all  the  points  of  the  plan, 
as  indicated  by  the  figures  and  letters,  and  from  these 
points  draw  lines  toward  the  center  As,  all  as  shown. 
From  A"  as  center,  with  radii  corresponding  to  the  dis- 
tances A  L,  A  2,  A  3,  A  4  and  A  M  of  the  elevation, 
strike  arcs  intersecting  corresponding  lines  just  drawn. 
Then  a  line  traced  through  the  intersections  thus  ob- 
tained will  be  the  shape  of  the  opening  to  be  cut  in  the 
envelope  of  the  cone  to  fit  the  end  of  the  cylinder. 


New 


Worker   Pattern  .AW-. 


PROBLEM   160. 

The  Patterns  of  a  Cylinder  Joining  the  Frustum  of  a  Cone  in  which  the  Axis  of  the  Cylinder  is 
Neither  at  Right  Angles  to  the  Axis  Nor  to  the  Side  of  the  Cone. 


The  principles  involved  in  the  solution  of  this 
problem  are  exactly  the  same  as  those  of  the  problem 
immediately  preceding,  to  which  the  reader  is  referred 
for  a  more  full  explanation  of  the  operation.  The  de- 


section  of  the  cylinder,  which  divide  into  anv  con- 
venient number  of  equal  pails,  as  indicated  b\-  the 
small  ligures,  and  from  these  points  carrv  lines  parallel 
with  X  X  cutting  the.  side  B  L  of  the  cone  in  the 


r.X' 


Fig.  SSI. — A  CyliniliT  J.iinnty  tin'  /'Viis/Hm  of  it  Cone  at  a.i   Oblique  Anvfle. 


tails  or  conditions  differ  only  in  the  angle  at  which  the 
cvlinder  joins  the  side  of  the  cone. 

In  Fig.  521,  let  C  B  L  be  the  elevation  of  a  right 
cone  of  which  C  c  I  L  is  a  frustum,  and  let  M  T  U  N 
represent  the  cylinder  which  is  to  join  the  frustum, 
making  the  angle  U  N  L  greater  than  a  right  anise. 
The  first  operation  will  be  to  determine  the  shape  of 
miter  line  M  X  of  side  elevation.  Draw  V  \V  X,  the 


points  A.  (!  and  K.  producing  them  until  they  cut  tin- 
axis  15  Y  in  the  points  11,  F  and  D.  Draw  a  plan  un- 
der the  side  elevation  of  cone,  as  shown  by  ()  S  Q  P, 
which  divide  into  any  convenient  number  of  equal 
parts.  From  points  1  to  4  in  S  Q  P  carry  lines  verti- 
cally to  the  base  C  L,  and  thence  toward  the  apex  B, 
cutting  the  lines  I)  K.  V  <i  and  II  A. 

Draw  a  second  elevation  of  the  cone,  as  shown  b\~ 


Pattern  Problems. 


299 


<'  l>  I,',  which  represents  the  cone  as  lurnc'il  quarter 
\v:iy  roiiml.  Draw  a  corresponding  plan  under  the  end 
elevation,  as  shown  bv  S'  <^'  I"  ()'.  Divide  this  plan 
into  the  same  number  <>f  equal  parts,  commencing  t<> 

number  them  at  the  sa point  as  in  the  other  plan — 

that  is.  at  the  point  Q.  From  the  points  1  to  4  in- 
elnsive  in  S'  </  P'  of  plan  carrv  vertical  lines  to  the 
base  ("  L'.  and  thence  to  the  apex  15". 

The  next  step  is  to  const  met  sections  of  the  com- 
as it  would  appear  if  cut  upon  the  planes  represented 
liy  the  lines  II  A.  KG  and  1)  K.  For  this  purpose 
place  the  "[-square,  at,  right  angles  to  the  axes  of  the 
two  cones,  and,  bringing  it  against  the  points  of  in- 
tersection of  the  lines  from  the  liase  C  L  with  D  E, 
cut  corresponding  lines  in  the  second  elevation,  and 


~     a  b 
Q 

t'iij.  ii2.'.—Enrfltipe  of  t'ouc  Shawn  in   t'iij.  SSI. 


through    the    points    of    intersection    thus    established 
trace  a  line,  as  shown  by  N'  K3  N*. 

Continue  the  axis  Y'  B"  as  may  be  convenient, 
upon  which  set  off  spaces  equal  to  those  between  the 
points  in  ]>  K,  and  through  these  points  draw  lines  at 
ri'jht  angles  to  ])'  K'.  Place  the  J-square  parallel  to 
the  axis  Y'  IV,  and,  bringing  it  against  the  several 
points  in  X'  K'  X'J,  cut  the  lines  drawn  through  D'  E1, 
a  <  shown,  and  through  these  points  of  intersection 
trace  u  line,  as  shown  by  X4  E'  W.  Then  N'  E1  N3  is 
a  section  of  the  cone  as  it  would  appear  if  cut  on  the 
line  1)  K.  Sections  corresponding  to  F  G  and  M  A 
can  be  obtained  in  a  similar  manner. 

Having  obtained  sections  of  the  cone   correspond- 
ing to  the  several  lines  D  E,  F  G  and  II  A,  it  will  next 


be  necessary  to  project  a  plan  at  right  angles  to  the 
axis  of  tie  cvlindcr.  in  which  each  of  these  section.- 
shall  lind  its  place.  Therefore-,  from  all  the  points  of 
the  cylinder  and  of  its  intersections  with  the  sides  and 
axis  of  the  cone  project  lines  at  right  angles  to  N  X 
indefinitely,  through  which  at  any  convenient  point, 
draw  a  line,  as  D"  X',  parallel  to  X  X.  I'poii  this  line, 
as  a  center  of  the  plan  about  to  be  constructed,  place 
the  oblique  sections  just  obtained  so  that  each  mav  be 
in  line  with  the  line  in  the  elevation  which  it  repre- 
sents, and  their  center  lines  shall  all  coincide  with 
D-  X1,  all  as  shown.  Make  TJ  17  equal  to  T  U  and 
complete  the  plan  of  the  cylinder,  opposite  the  end 
of  which  draw  a  prolilc,  as  indicated  bv  V  W  X', 
commencing  the  divisions  at  the  point  V.  From  the 
several  points  in  the  profile  V  W  X'  drop  lines  paral- 
lel with  the  center  line  D2  X1  against  the  several 
proiiles  ,/  Iv  d',f  G*  f  and  /*  A2  /<',  and  thence  drop 
the  points  back  on  the  elevation,  cutting  correspond- 
ing lines  in  it.  Thus,  from  the  intersection  of  the 
line  drawn  from  point  W  (3)  with  G2/'1  of  section  cut 
the  lino  F  G,  which  in  the  elevation  corresponds  to 
the  point  3  in  the  profile  V  W  X.  From  the  intersec- 
tion of  a  line  drawn  from  point  '2  in  V  W  with'A'J  //' 
of  section  cut  the  line  II  A,  and  so  on,  as  indicated 
by  the  dotted  lines.  A  line  traced  through  these 
points  of  intersection,  as  shown  by  the  curved  line 
M  N,  will  be  the  miter  line  in  elevation,  from  which 
the  patterns  can  be  obtained  as  follows: 

For  the  pattern  of  cylinder  shown  in  elevation  by 
M  T  U  N,  on  T  U  extended  lay  off  a  stretchout  of 
profile  V  W  X,  through  the  points  in  which  draw  the 
usual  measuring  lines.  Place  the  J-square  parallel 
with  T  U,  and,  bringing  it  against  the  points  in  the 
miter  line  M  N,  cut  measuring  lines  of  corresponding 
number.  Trace  a  line  through  the  points  thus  ob- 
tained, as  shown  from  m  to  ?/*'.  Then  in  /  f  ni  is  the 
pattern  of  the  cylinder  to  lit  against  the  cone,  as  shown 
in  elevation  by  M  T  U  N. 

To  obtain  the  pattern  of  the  frustum  carrv  lines 
from  each  of  the  points  in  the  miter  line  M  X  liomini-* 
tallv  across  the  elevation,  cutting  the  side  of  the  frus- 
tum c  C,  as  shown  by  a1,  I1  and  d' ;  also  through  the 
same  points  draw  lines  from  the  apex  13.  cutting  the 
base  line  C  L,  and  thence  drop  them  on  the  plan,  as 
sho.vn  by  1,  ~2,  a  and  b.  From  any  convenient  point, 
as  IV  in  Fig.  ~>-l-2,  as  a  center,  with  radii  equal  to  B  c 
and  1>  C.  describe  arcs,  as  shown  by  O  Q  O'  and  c1  /'. 
M  ike  O  Q  O1  equal  in  length  to  the  plan  of  cone  0  S 
Q  P  and.  upon  it  set  off  e-ich  way  from  the  point  Q 


300 


Tlie  New  Metal   Worker  Pattern  Book. 


spaces  equal  to  those  upon  the  plan  between  Q  and  S. 
From  these  points  draw  lines  indefinitely  toward  the 
center  B5.  With  B"  as  center  describe  arcs  whoso 
radii  correspond  to  B  M,  B  a1,  B  b',  B  d1  and  B  N",  cutting 
lines  of  corresponding  number  or  letter.  Then  a  line 


traced  through  the  intersections  thus  obtained  will  be 
the  shape  of  opening  to  cut  in  the  envelope  of  frustum 
where  it  joins  the  cylinder,  and  lines  drawn  from  0 
and  0'  toward  B2  till  they  cut  the  arc  c'  I'  in  the  points 
c'  and  I1  will  complete  the  pattern  of  the  frustum. 


PROBLEM   161. 

The  Patterns  of  Two  Cones  of  Unequal  Diameter  Intersecting;  at  Right  Angles  to  their  Axes. 


Let  U  T  V  in  Fig.  523  be  the  elevation  of  a  cone, 
at  right  angles  to  the  axis  of  which  another  cone  or 
frustum  of  a  cone,  0  F  G  P,  is  to  miter.  Let  L  K  N  M 
be  a  section  of  the  frustum  on  the  line  F  G.  Let  U" 
W  Vs  be  a  half  plan  of  the  larger  cone  at  the  base. 
The  first  step  in  describing  the  patterns  is  to  obtain  the 
miter  line  in  the  elevation,  as  shown  by  the  curved  line 
from  0  to  P.  With  this  obtained  the  development  of 
the  pattern  is  a  comparatively  simple  operation. 

To  obtain  the  miter  line  O  P  proceed  as  follows : 
Divide  the  profile  L  K  N  M  into  any  convenient  num- 
ber of  equal  parts,  as  shown  by  the  small  figures.  In- 
asmuch as  the  divisions  of  this  profile  are  used  in  the 
construction  of  the  sections — or,  in  other  words,  since 
sections  through  the  cone  must  be  constructed  to  cor- 
respond to  certain  lines  through  this  profile — it  is  de- 
sirable that  each  half  be  divided  into  the  same  number 
of  equal  parts,  as  shown  in  the  diagrams.  Thus  2  and 
2,  3  and  3,  4  and  4  of  the  opposite  sides  correspond, 
and  sections,  shown  in  the  upper  part  of  the  diagram, 
are  taken  upon  the  planes  which  they  represent.  From 
the  points  in  the  profile  L  K  N  M  draw  lines  parallel 
to  B  E  cutting  the  end  F  G  of  the  frustum.  Produce 
the  sides  0  F  and  P  G  until  they  meet  in  E,  which  is 
the  apex  of  the  cone.  Through  the  points  in  F  G  draw 
lines  from  E,  producing  them  until  they  cut  the  axis 
of  the  cone,  as  shown  at  A,  A1,  A". 

Next  construct  sections  of  the  cone  as  it  would 
appear  if  cut  through  upon  the  lines  A  C,  A1  B,  A'  D. 
Divide  the  plan  U"  W  V"  into  any  convenient  number 
of  parts.  From  the  points  thus  established  carry  lines 
vertically  to  the  base  line  U  V,  and  thence  carry  them 
toward  the  apex  T,  cutting  the  lines  A  C,  A1  B,  A"  D, 
all  as  shown.  Through  each  of  the  several  points  of 
intersection  in  these  lines  draw  horizontal  lines  from 
the  axis  of  the  cone  to  the  side,  all  as  shown.  At  right 
angles  to  the  lines  A  C,  A1  B,  A2  D  project  lines  to 
any  convenient  point  at  which  to  construct  the  re- 
quired sections.  Upon  the  lines  drawn  from  the  points 


A,  A1,  A"  locate  at  convenience  the  points  A3,  A4, 
A".  Inasmuch  as  A1  B  is  at  right  angles  to  the  axis 
of  the  cone,  the  section  corresponding  to  it  will  be  a 
semicircle  whose  radius  will  be  equal  to  A1  B.  There- 
fore, from  A3  as  center,  with  radius  A1  B,  describe  the 
semicircle  S  B1  R.  For  the  section  corresponding  to 
A2  D  lay  off  from  A6  the  distances  A'  S1  and  A'  R",  in 
a  line  drawn  at  right  angles  to  A"  D  of  the  elevation, 
each  in  length  equal  to  the  horizontal  line  drawn 
through  the  point  A"  from  the  axis  to  the  side  of  the 
cone.  At  right  angles  to  S"  Ra  draw  A'  D1,  in  length 
equal  to  A"  D  of  the  elevation.  Set  off  in  it  points  5, 
4  and  2,  corresponding  to  similar  points  in  A"  D  of  the 
elevation.  Through  these  points  5,  4:  and  2,  at  right 
angles  to  A6  D1,  draw  lines  indefinitely.  From  A5  as 
center,  with  radius  equal  to  the  length  of  horizontal 
line  passed  through  point  5  in  A"  D  of  the  elevation, 
describe  an  arc  cutting  line  5  of  the  section.  From  the 
same  center,  with  a  radius  equal  to  the  length  of  the 
horizontal  line  drawn  through  point  4  in  the  line  A"  1 ) 
of  the  elevation,  strike  an  arc  cutting  the  line  4,  etc. 
Then  a  line  traced  through  these  points,  as  shown  by 
S3  D'  R",  will  be  the  section  of  the  cone  as  it  would 
appear  if  cut  on  the  line  A"  D  of  the  elevation.  In  like 
manner  obtain  the  section  S'  C'  R1,  corresponding  to 
A  C  of  the  elevation. 

These  sections  may,  if  preferred,  be  obtained  in 
ths  manner  described  in  connection  with  Problems  159 
and  160. 

As  these  sections  are  obtained  solely  for  the  pur- 
pose of  determining  at  what  point  in  their  perimeters — 
that  is,  at  what  distance  from  points  C',  B1  and  D' — they 
will  be  intersected  by  the  lines  representing  the  points 
2,  3  and  4  of  the  profile  L  K  M  N,  it  is  not  necessary 
that  the  complete  half  sections  should  be  developed. 
In  the  engraving,  the  small  intersecting  cone  has  so 
little  flare  that  the  lines  A  C  and  A3  D  cross  the  large 
cone  so  nearly  at  right  angles  to  its  axis  that  sections 
2  2  and  4  4  could  be  constructed  with  sufficient  accu- 


Pattern  Problems. 


301 


4-. 


P2 


.  523.— The  Patterns  of  Two  Unequal  Cones  Intersecting  at  liirjht  Angles  tu  their  Axes. 


302 


77"    New    .!/</"/    Wurtcer  Pattern  Book. 


racy  for  practical  purposes,  as  in  the  ca.-e  of  section  :;  .".. 
bv  small  arcs  of  circles  with  radii. respectively  equal  to 
A  C  and  A*  D,  and  of  only  sufficient  length  to  include 
tUc  points  c  c  and  d  d. 

Prolong  A'  D1,  as  shown  by  E1,  making  A:  K'  in 
length  equal  to  A1  E  of  the  elevation.  In  like  manner 
make  A'  E'  and  A3  E1  equal  to  A  E  and  A1  E  of  the 
elevation  respectively.  At  right  angles  to  these  lines 
in  the  sections  set  off  F*  G1,  F'  G',  F'  G4,  in  position 
eoi responding  to  F  G  of  the  elevation.  Mike  the 
length  of  F*  G'  equal  to  the  length  across  the  section 
of  the  frustum  marked  'J  2.  In  like  manner  make  Fa 
G3  equal  to  3  3,  and  F'  G'  equal  to  i  4  of  the  section. 
From  E',  E'  and  E3  respectively,  through  these  points 
in  the  several  sections,  draw  lines  cutting  the  oblique 
sections  just  obtained.  From  the  several  points  of  in- 
tersection between  the  lines  drawn  from  E',  E2,  E3  and 
the  sections  of  the  cone,  as  shown  by  d  <1,  c  c  and  I>  b, 
carry  lines  back  to  the  elevation,  intersecting  the  lines 
A  C,  A1  B,  A2  D.  Then  a  line  traced  through  these 
several  intersections,  as  shown  from  O  to  P,  will  be 
the  miter  line  in  elevation. 

Having  thus  obtained  the  miter  line,  proceed  to 
describe  the  patterns,  as  follows  :  For  the  envelope  of 
the  small  cone,  from  E  as  center,  with  radius  E  G,  de- 
scribe the  arc  F1  G1,  upon  which  set  off  the  stretchout 
of  the  section  L  K  M  X.  Through  the  points  in  this 
arc,  from  E,  draw  radial  lines  indefinitely.  From  E  as 
center,  with  radii  corresponding  to  the  several  points 
in  the  miter  line  0  P,  but  obtained  from  the  oblique 


sections  above,  cut  corresponding  radial  lines.  Thus 
with  the  radius  E11  «/  cut  lines  -i  and  -t,  with  the  radius 
Iv  '•  cut  lines  i'  and  -2  and  with  radius  E1  b  cut  lines  :! 
and  3. 

Then  a  line  traced  through  these  points  of  intersec- 
tion, as  shown  by  P'  ( )'  P-',  will  be  the  shape  of  the 
pattern  of  the  frustum  1"  lit  against  the  larger  cone. 

For  the  pattern  of  the  larger  cone,  from  T  us 
eenter,with  radius  T  I",  describe  the  are  V  U',  in  length 
equal  to  the  circumference  of  the  entire  plan  of  the 
cone.  From  the  points  in  the  miter  line  O  P  earrv 
lines  parallel  to  the  base  of  the  cone  cutting  its  side 
T  U,  as  shown  between  ()3and  P',  also  through  the 
points  in  O  P  draw  lines  from  the  apex  cutting  the  base 
and  thence  earrv  them  \crticallv  to  tin1  plan. 

These  points  can  be  numbered  upon  the  side  of 
the  cone  to  correspond  with  the  plan,  but  entirely  in- 
dependent of  the  system  of  numbers  emplnved  upon 
the  smaller  cone.  Upon  the  arc  V  U'  set  off  points 
corresponding  to  the  points  just  obtained  in  the  plan 
from  the  miter  line,  from  which  draw  lines  toward  the 
center  T.  With  one  foot  of  the  compasses  set  at  the 
point  T,  bring  the  pencil  point  successively  to  the 
points  between  O3  and  P3  and  cut  radial  lines  of  corre- 
sponding number  in  the  pattern.  Then  a  line  traced 
through  these  intersections,  as  shown  by  X  V  Z  V. 
will  be  the  shape  of  the  opening  to  be  cut  in  the  en- 
velope of  the  larger  cone,  over  which  the  smaller  cone 
will  fit,  and  T  U'  A*'  will  be  the  envelope  of  the  entire 
cone. 


PROBLEM    162. 


The  Patterns  of  the  Frustums  of  Two  Cones  of  Unequal  Diameters  Intersecting:  Obliquely. 


In  Fig.  524,  let  M  X  P  O  be  the  side  elevation  of 
the  larger  frustum  and  F'  G'  S  R  the  side  elevation 
of  the  smaller,  the  two  joining  upon  a  line  between 
the  points  R  and  S,  which  line  must  be  obtained  be- 
fore the  patterns  can  be  developed.  Produce  the  sides 
S  G1  and  R  F'  until  they  meet  in  the  point  E.  At  any 
convenient  place  on  the  line  of  the  axis  of  the  smaller 
frustum  draw  the  profile  II  F  K  G,  corresponding  to 
the  end  F1  G1.  Divide  this  profile  into  any  convenient 
number  of  equal  parts,  as  shown  by  the  small  figures 
1,  2,  3,  etc.,  and  from  these  divisions,  parallel  to  the 
axis  of  the  cone,  drop  points  on  to  F'  G'.  From  the 
apex  E,  through  these  points  in  F1  G1,  carry  lines 


cutting  the  side  F  P  of  the  larger  frustum,  and  pro- 
ducing them  until  they  meet  the  center  line,  or  the 
base  O  P,  all  as  shown  by  1?  A.  C  A1  and  D  A". 

The  next  step  is  to  construct  sections  of  the  larger 
frustum  as  it  would  appear  if  cut  on  each  of  these 
lines,  from  which  to  obtain  points  of  intersection  with 
the  lines  of  the  smaller  frustum  for  determining  the 
miter  line  from  R  to  S  in  the  elevation.  Draw  the 
plan  of  the  base  of  the  larger  frustum,  as  shown  bv 
T  U  V  \V.  and  divide  one-half  of  it  in  the  usual  man- 
ner. From  these  points  carry  lines  vertically  to  the 
base  ()  P  of  the  frustum.  Produce  the  sides  ()  M  and 
P  X  until  they  meet  in  the  point  L.  From  the  points 


Pattern  Problems. 


303 


in   the   base  line  obtained   from   the  plan  carry  lines      project  lines   at  right  angles  cutting  A"  B1,  as  shown, 
toward  the  apex  L,    cutting  the   section   lines  A  B,      in  the  points  4,  3  and  2.     In  like  manner  make  A4  C1 


. _ -ji-  ^ 


O'ff  'TI 


Fig.  514' — Patterns  of  the  Frustums  of  Two  Unequal  Cones  Intersecting  Obliquely. 


A'  C  and  A3  D,  as  shown.  Parallel  to  A  B  and  of  the 
same  length,  at  any  convenient  point  outside  of  the 
elevation,  draw  A3  B1,  and  from  the  points  in  A  B 


equal  and  parallel  to  A  C,  and  from  the  points  in  A  C 
project  lines  at  right  angles  to  it,  cutting  it  as  shown, 
giving  the  points  4,  3  and  2,  Also  make  A'  P'  equal 


304 


The  New  Metal   Worker  PMern  Book. 


to  the  section  line  A'  D  of  the  elevation,  and  cut  it  by 
lines  from  the  points  in  A"  D,  obtaining  the  points 
3  and  2,  as  shown.  In  order  to  complete  these  sev- 
eral sections,  the  width  of  the  frustum  through  each 
of  the  points  indicated  is  to  be  set  off  on  correspond- 
ing lines  drawn  through  A3  B1,  A4  C1  and  A6  D1.  To 
obtain  the  width  through  these  points  first  draw  an 
end  elevation  of  the  larger  frustum,  as  shown  by  M1  N' 
P'  0'.  Produce  the  sides,  obtaining  the  apex  L'. 
Draw  a  plan  and  divide  it  into  the  same  number  of 
spaces  as  that  shown  in  T  U  V  W,  and  commence 
numbering  at  a  corresponding  point,  all  as  indicated 
by  V  IP  T1  W.  From  the  points  in  the  plan  carry 
lines  vertically  to  the  base  0'  P1,  and  thence  toward 
the  apex  L1.  Place  the  blade  of  the  T-square  at  right 
angles  to  the  axis  of  the  cone,  and,  bringing  it  succes- 
sively against  the  points  in  the  section  line  A  B  in  the 
side  elevation,  draw  lines  cutting  the  axis  of  the  end 
elevation,  and  cutting  the  lines  corresponding  in  num- 
ber to  the  several  points  in  A  B,  all  as  shown  by  a  a, 
b  b  and  c  c.  Make  the  length  of  the  lines  drawn 
through  A3  B'  equal  to  the  corresponding  lines  thus 
obtained,  as  shown  by  a1  a1,  b1  b\  c1  c1  and  d1  d\  and 
through  these  extremities  trace  a  line,  as  shown  by 
d'  B1  d',  which  will  be  the  section  through  the  cone 
when  cut  on  the  line  A  B.  In  like  manner  complete 
the  sections  P  C1  I1  and  /'  D'  /'. 

As  remarked  in  the  previous  problem,  it  is  only 
necessary  that  these  sections  should  be  developed  far 
enough  from  the  points  B',  C1  and  D'  to  receive  the 
lines  representing  the  sections  of  the  smaller  frustum. 
Produce  A8  B1,  making  B'  E1  equal  to  B  E  of  the  ele- 
vation, and  B'  X3  equal  to  B  X'  of  the  elevation.  In 
like  manner  make  C1  E3  equal  to  C  E,  and  C'  X1  equal 
to  G  X.  Make  D1  E3  equal  to  D  E,  and  D1  X'  equal 
to  D  X3.  Through  X3,  at  right  angles  to  B'  E',  draw 
a  line  in  length  equal  to  the  line  2  2  drawn  across  the 
profile  F  K  G  II,  with  which  this  section  corresponds, 
as  shown  by  2'  2'.  Through  X4  draw  a  line  equal  to 
II  K,  as  shown  by  H1  K',  and  through  Xs  draw  4'  4', 
in  length  equal  to  the  line  4  4  drawn  through  the 
profile  F  K  G  H.  From  E',  through  the  extremities 
of  2'  2',  draw  lines  cutting  the  section.  In  like  man- 
ner draw  lines  from  E3  through  the  points  H'  K1,  and 
from  E3  through  the  points  4'  4'.  From  the  points  at 
which  these  lines  meet  the  sections,  as  a"  a'  in  the  first, 
o  o  in  the  second  and  m  m  in  the  third,  carry  lines 
back  at  right  angles  to  and  cutting  the  corresponding 
section  lines  in  the  elevation.  A  line  traced  through 
the  points  thus  obtained,  as  shown  by  E  S,  is  the 


miter  line  in  elevation  formed  by  the  junction  of  the 
two  frustums. 

Having  thus  obtained  the  miter  line  in  elevation, 
proceed  to  develop  the  patterns  as  follows :  From  the 
points  in  B  S,  at  right  angles  to  A1  E,  which  is  the 
axis  of  the  smaller  cone,  draw  lines  cutting  the  side 
K  S,  as  shown  by  the  small  figures  1,  2,  3,  4  and  o. 
These  points  are  to  be  used  in  laying  off  the  pattern 
of  the  smaller  frustum.  From  E  as  center,  with 
radius  E  G1,  describe  the  arc  F3  G",  upon  which  step 
off  the  stretchout  of  the  profile  F  K  H  G,  numbering 


Fig.  525. — The  Envelope  of  the  Larger  Frustum  Shoivn  in  Fig.  524. 

the  points  in  the  usual  manner.  Through  the  points, 
from  the  center  E,  draw  radial  lines  indefinitely. 
From  the  same  center,  E,  with  radius  E  1  (of  the  . 
points  in  E  S),  cut  the  radial  line  numbered  1,  and  in 
like  manner,  with  radii  E2,  E3,  etc.,  cut  the  correspond- 
ing numbers  of  the  radial  lines.  A  line,  E1  S1,  traced 
through  the  several  points  of  intersection  thus  formed 
will  be  the  larger  end  of  the  pattern  for  the  small  frus- 
tum, thus  completing  the  shape  of  that  piece,  all  as. 
shown  by  E1  S1  G' F3. 


Patient   Problems. 


305 


To  avoid  confusion  of  lines,  the  manner  of  ob- 
taining the  envelope  of  the  large  frustum  is  shown  in 
Fig.  ">-•">,  which  is  a  duplicate  of  the  side  elevation 
and  plan  shown  in  Fig.  r>i!-t,  the  miter  line  R1  S'  and 
the  points  in  it  being  the  same.  Similar  letters  refer 
to  corresponding  parts  in  the  several  figures.  From 
L*  as  center,  with  radius  L"  O'1,  describe  an  arc,  as 
shown  by  V  Z,  and  from  the  same  center,  with  radius 
I/'  M",  describe  a  second  arc,  as  shown  by  y  z.  Draw 
V  >/.  and  upon  V  Z  lav  off  the  stretchout  of  the  plan 
V'WT,  all  as  shown.  Draw  Z  z.  Then  ZzyY 
will  be  the  envelope  of  the  large  frustum.  Through 
th?  points  in  the  miter  line  R'  S'  draw  lines  from  the 
apex  of  the  cone  to  the  base,  and  from  the  base  con- 
tinue them  at  right  angles  to  it  until  they  meet  the 


circumference  of  the  plan.  Mark  corresponding  points 
in  the  stretchout  Y  Z,  and  insert  any  points  which  do 
not  correspond  with  points  already  fixed  therein. 
From  eacli  of  the  points  thus  designated  draw  a  line 
across  the  envelope  already  described  to  the  apex,  as 
shown  by  3  La,  x  L"  and  1  L'.  Also,  from  the  points 
in  the  miter  line  H'  S1  draw  lines  at  right  angles  to  the 
axis  of  the  frustum  cutting  the  side  La  0",  as  shown. 
From  L"  as  center  describe  arcs  corresponding  to  each 
of  these  points  and  cutting  the  radial  lines  drawn 
across  the  envelope  of  the  cone.  A  line  traced  through 
the  points  of  intersection  between  arcs  and  lines  of 
the  same  number,  as  shown  by  h  R1  h'  S2,  will  be  the 
shape  of  the  opening  to  fit  the  base  of  the  smaller 
frustum. 


306 


The  Xciv  Metal    \\'orkcr  PiUttm  Buuk. 


SECTION  3. 


(TRIANGULATION.) 


The  class  of  subjects  treated  in  this  section  will 
include  all  irregular  forms  which  can  be  constructed 
from  sheet  metal  by  simple  bending  or  forming,  but 
whose  patterns  cannot  be  developed  by  the  regular 
methods  employed  in  the  two  previous  sections  of  this 
chapter.  These  problems  divide  themselves  naturally, 
in  regard  to  the  arrangement  of  the  triangles  used  in 
the  development  of  the  patterns,  into  two  classes,  viz.  : 
First,  those  in  which  the  vertices  of  the  triangles  used 
in  constructing  the  envelopes  all  terminate  at  a  com- 
mon point  or  apex,  and,  second,  those  in  which  the 
relative  position  of  the  base  and  the  vertex  is  reversed 
in  each  succeeding  triangle,  or,  in  other  words,  those 
in  which  the  vertices  of  alternate  triangles  point  in 
opposite  directions. 

In  the  introduction  to  Section  2  (page  2-tO),  at- 
tention is  called  to  the  difference  between  a  scalene  or 
oblique  cone  and  a  right  cone  with  an  oblique  base. 
The  scalene  cone  may  be  called  the  type  or  representa- 
tive of  a  large  number  of  forms  belonging  to  the  first 


elass  above  mentioned,  since  many  rounded  surfaces 
entering  into  the  construction  of  various  irregular  liar- 
ing  articles  are  portions  of  the  envelope  of  a  scalene 
cone.  The  principles  involved  in  this  particular  class 
of  forms  are  explained  in  that  part  of  Chapter  V  refer- 
ring to  Fig.  271,  page  it-i. 

Inasmuch  as  triangulation  is  resorted  to  in  all  cases 
\vliere  regular  methods  are  not  applicable,  it  is  not  sur- 
prising that  the  forms  here  treated,  especially  those 
included  in  the  second  elass  above  referred  to,  are  more 
varied  in  character  than  those  of  anv  other  class  to  be 
met  with  in  pattern  cutting.  An  explanation  of  methods 
and  principles  governing  these  will  be  found  in  the 
third  subdivision  of  Chapter  Y,  beginning  on  page  86. 
The  last  few  problems  of  this  class  are  devoted  to  the 
development  of  the  horizontal  surfaces  of  arches  in  cir- 
cular walls. 

The  arrangement  of  problems  in  this  section  will  be 
in  accordance  with  the  above  classification  although  no 
headings  will  be  introduced  to  distinguish  the  classes. 


PROBLEM   163. 

The  Envelope  of  a  Scalene  Cone. 


In  Figs.  526  and  527  are  shown  perspective  rep- 
resentations of  scalene  or  oblique  cones.  In  Fig.  .")2i> 
the  inclination  of  the  axis  to  the  base  is  so  great  that 
a  vertical  line  dropped  from  its  apex  would  fall  out- 
side the  base,  while  in  Fig.  527  a  perpendicular  from 
its  apex  would  fall  at  a  point  between  the  center  and 
the  perimeter  of  its  base. 

Supposing  the  circumference  of  the  base  in  either 
case  to  be  divided  into  a  number  of  equal  spaces,  it  is 
plain  to  be  seen  that  lines  drawn  upon  the  surface 
of  the  cone  from  the  points  of  division  to  the  apex 
would  be  straight  lines  of  unequal  lengths,  and  that 
such  lines  would  divide  the  surface  of  the  cone  into 
triangles  whose  vertices  are  at  the  apex  of  the  cone 
and  whose  bases  would  be  the  divisions  upon  the  base 
of  the  cone.  It  will  be  seen  further  that  with  the 
means  at  hand  of  determining  the  lengths  of  these 
lines  forming  the  sides  of  the  triangles,  the  pattern 


cutter  possesses   all    that  is  necessary   in   developing 
their  envelopes  or  patterns. 


Fig.  'CM. 

Scalene  Cones  of  Different 


Fig.  &!7. 


In   Fig.  52S,  D  A  II  is  an   elevation   of  the   cone 
shown  in  Fig.  526  and  D  G  H  is  a  half  plan  of  the 


Pattern  />/ •«/</<  ms. 


307 


same,  drawn  for  convenience,  so  tliat  I)  II  is  at  <>nce 
the  base  line  of  the  elevation  and  the  eenter  line  of 
the  plan.  Fig.  .~>'2\l  shows  an  elevation  and  plan  of  the 
eoiio  shown  in  Fig.  ->'1~ .  drawn  in  the  same  manner. 
The  principle  involved  in  the  development  of  the  pat- 
terns of  the  two  oblique  cones  is  exactly  the  same  and, 
as  will  be  seen,  letters  referring  to  similar  parts  in  tin- 
two  drawings  are  the  same;  therefore  the  following 
demonstration  will  applv  equally  well  in  either  ease. 
From  the  apex  A  drop  a  perpendicular  to  the  base 


Fiij.  MS.— Pattern  of  Cone  Shown  in  Fig.    ~>M. 

line,  locating  the  point  N".  Divide  the  base  D  <r  11 
into  any  convenient  number  of  equal  spaces,  as  shown 
by  the  small  figures,  and  from  the  points  thus  ob- 
tained draw  lines  to  the  point  N.  These  lines  will 
form  the  bases  of  a  series  of  right  angled  triangles  of 
which  A  N"  is  the  perpendicular  night,  and  whose  hv- 
po  then  uses  when  drawn  will  give  the  correct  length  of 
lines  extending  from  the  points  of  division  in  the  base 
of  the  cone  to  the  apex.  The  most  convenient  method 
of  constructing  these  right  angled  triangles  is  to  trans- 
fer the  distances  from  N  to  the  various  points  upon 


the  circumference  of  the  base  to  the  line  X  I)  as  a  base 
line,  measuring  each  time  from  the  point  X,  by  which 
method  the  line  A  X  becomes  the  common  perpen- 
dicular of  all  the  triangles.  Therefore  from  N  as 
center,  with  the  distances  N  1,  N  2,  etc.,  as  radii,  de- 
scribe ares  as  shown  in  the  engraving,  cutting  the  base 
I  iiu.'  N  D.  Lines  from  each  of  these  points  to  the 
apex,  as  A  1,  A  2,  etc.,  will  be  the  required  hypoth- 
enuses. 

The  simplest  method  of  developing  the  pattern  is 
to  iirst  describe  a  number  of  arcs  whose  radii  are  re- 
spectively equal  to  the  various  hypothenuses  just 
obtained  ;  therefore  place  one  foot  of  the  compasses  at 
A.  and,  bringing  the  pencil  point  successively  to  the 


Fig.  53).— Pattern  <>/  Cone  Shown  in  J-'ii/.  527. 

points  1,  2,  3,  etc.,  upon  the  line  N  D,  describe  arcs 
indefinitelv.  From  any  point  upon  the  arc  drawn  from 
point  1,  as  ->i ,  draw  a  line  to  A  as  one  side  of  the  pat- 
tern. Next  take  between  the  feet  of  the  dividers  a 
space  equal  to  the  spaces  upon  the  circumference  of 
the  plan,  and  placing  one  foot  of  the  dividers  at  the 
point  »,  swing  the  other  foot  around  till  it  cuts  the 
arc  drawn  from  point  2 ;  then  A  »  ~2  will  be  the  first 
triangle  forming  part  of  the  envelope  or  pattern. 
\Vith  the  same  space  between  the  points  of  the  dividers, 
and  2  of  the  pattern  as  center,  swing  the  dividers 
!  around  again,  cutting  the  arc  drawn  from  point  3. 
Repeat  this  operation  from  3  as  center,  or,  in  other 
words,  continue  to  step  from  one  are  to  the  next,  until 
all  the  ares  have  been  reached,  as  at  '/.  which  in  this 


308 


The  New  Metal   Worker  Pattern  Book. 


case  will  constitute  one-half  the  pattern ;  after  which, 
if  desirable,  the  operation  of  stepping  from  arc  to  arc 
may  be  continued,  as  shown,  finally  reaching  the  point 


d.  Draw  d  A  and  trace  a  line  through  the  points  ob- 
tained upon  the  arcs,  as  shown  by  ;i  <j  d,  which  will 
complete  the  pattern. 


PROBLEM   164. 

The  Envelope  of  an  Elliptical  Cone. 


In  Fig.  530  is  shown  an  elevation  and  plan  of  a 
cone  whose  base  is  an  elliptical  figure.  So  far  as  the 
solution  of  this  problem  is  concerned  the  plan  may  be 
a  perfect  ellipse  or  an  approximate  ellipse  drawn  by 
any  convenient  method.  Fig.  531  shows  a  perspective 
view  of  the  cone  in  question,  upon  which  lines  have 
been  drawn  from  points  assumed  in  its  base  to  the 


Fig.  530.— "Plan  and  Elevation  of  Cone  with  Elliptical  Base. 

apex.  From  an  inspection  of  this  view  it  will  appear 
that  these  lines,  as  in  the  case  of  the  scalene  cone,  are 
of  unequal  length,  and  therefore  that  the  pattern  of  its 
envelope  may  be  developed  by  a  method  analogous  to 
that  adopted  in  the  preceding  problem. 

Since  the  cone  consists  of  four  symmetrical  quar- 
ters, it  will  be  necessary  to  obtain  the  envelope  of  only- 
one  quarter,  from  which  the  remainder  of  the  pattern 
can  be  obtained  by  reduplication.  Therefore  draw  a 
half  side  elevation,  as  shown  by  Y  X  C  of  Fig.  532, 
immediately  below  which  draw  a  quarter  plan,  X  C  E, 


so  that  the  line  X  C  shall  be  common  to  both  views,  as 
shown.     Divide  E  C  into  any  convenient  number  of 


Fig.  531. — ferspective  View  of  Cone  Shown  in  Fig.  530  with  Lines 
Drawn  upon  its  Surface  Used  in  Developing  Pattern. 

equal  parts,  as  indicated  by  the  small  figures.  Lines 
drawn  from  the  points  in  E  C  to  X  will  give  the  base 
lines  of  a  set  of  triangles,  whose  altitudes  are  equal  to 


QUARTER  PLAN 
Fig.  532.— Pattern  of  One-Half  the  Cone  Shown  in  Fig.  530. 

the  hight  of  the  article  X  Y,  and  whose  hypothenuses 
will  give  the  true  distances  from  the  apex  to  the  points 


Pattern  Problems. 


309 


assumed  in  the  base  line.  A  convenient  method  for 
drawing  these  triangles  is  as  follows:  With  X  as  cen- 
ter strike  arcs  from  the  points  in  E  C,  cutting  X  C,  as 
shown  by  the  small  figures.  Linos  drawn  from  the 
points  thus  obtained  to  Y,  as  shown,  will  give  the  hy- 
pothenuses  of  the  triangles.  With  Y  as  center,  and  the 
distance  from  Y  to  the  several  points  in  X  C  as  radii, 
strike  arcs  indefinitely.  From  Y  to  any  point  upon  the 
arc  0  draw  any  line,  as  Y  E,  which  will  form  the  edge 
of  pattern  corresponding  to  X  E  of  plan.  With  the 


dividers  set  to  the  space  used  in  stepping  off  E  C  of 
plan,  and  starting  from  E  of  the  pattern,  space  off  the 
stretchout  of  the  plan,  stepping  from  one  arc  to  the 
next,  as  shown.  From  the  point  6,  or  C',  draw  a  line 
to  Y.  Through  the.  points  thus  obtained  trace  a  line. 
Then  Y  C  E  is  the  pattern  for  that  part  of  the  article 
shown  on  plan  by  E  X  C.  This  quarter  can  be  du- 
plicated by  any  means  most  convenient  so  as  to  obtain 
the  pattern  for  one-half  or  for  the  whole  envelope  in 
one  piece,  as  desired. 


PROBLEM    165. 

Pattern  for  a  Raised  Boiler  Cover  With  Rounded  Corners. 


The  shape  of  the  cover  considered  in  this  problem 
may  perhaps  be  more  accurately  described  as  that  of 


END 


ELEVATION 


K.  J 

Fig.  533. — Plan  and  Elevation  of  Cover. 


I      2 
H 


an  oblong  pyramid  with  rounded  corners,  as  shown  by 
the  plan  and  elevation  in  Fig.  533,  an  inspection  of 


which  will  also  show  that  the  rounded  corners  are  por- 
tions of  a  scalene  cone,  while  the  four  pyramidal  sides 
are  simply  plain  triangular  surfaces. 

The  plan  shows  one-half  of  cover,  or  as  much  as 
would  usually  be  made  from  one  piece.  First  divide 
G  H  of  plan  into  any  convenient  number  of  equal 
parts — in  this  case  four — and  connect  the  points  thus 
obtained  with  0,  thus  obtaining  the  base  lines  of  a  set 
of  right  angled  triangles  whose  hypothenuses  when 
obtained  will  give  the  true  distances  from  the  points  in 
G  H  to  the  apex  of  the  cover. 

To  construct  a  diagram  of  triangles  represented 
by  lines  in  plan,  draw  the  right  angle  M  N  P  in  Fig. 


Fig.  BS4. — Diagram  of  Triangles. 

534,  making  M  N  equal  to  the  hight  of  cover,  as 
shown  by  B  D  of  elevation.  Measuring  from  N,  set 
off  on  N  P  the  length  of  lines  in  plan,  including  J  O 
and  O  F.  From  the  -points  in  N  P  draw  lines  to  M, 
as  shown.  The  line  C'  M  -gives  the  slant  hight  of 
cover  as  seen  in  the  end  elevation,  and  M  J'  the  slant 
hight  as  would  be  seen  in  side  elevation.  The  other 
lines  give  the  hypothenuses  of  triangles,  the  bases  of 
which  are  shown  by  the  lines  in  0  G  II  of  plan. 

To    describe   the    pattern    proceed    as    follows : 


310 


Tlie  New  Metal   Worker  Pattern  Book. 


Draw  the  line  Q  U,  in  Fig.  535,  in  length  equal  to 
M  J'  in  the  diagram  of  triangles.  Through  U,  at  right 
angles  to  Q  U,  draw  V  T,  making  U  T  and  U  V  each 
equal  to  J  II  or  J  K  of  Fig.  533,  and  draw  Q  T  and 
Q  V.  Then  Q  V  T  will  be  the  pattern  of  one  of  the 
sides  of  the  pyramid,  to  which  may  be  added  on  either 
side  the  envelope  of  the  portion  of  a  scalene  cone  shown 
by  H  0  G  in  Fig.  533.  It  should  be  here  remarked 
that  the  method  above  employed  of  obtaining  the 
length  Q  T  produces  the  same  results  as  that  employed 
in  the  diagram  of  triangles  as  shown  by  the  hypoth- 
enuse  M  1,  which  is  one  side  of  the  adjacent  triangle 
forming  part  of  the  pattern  of  the  rounded  corner.  From 
Q  of  Fig.  535  as  center  and  M  2  of  the  diagram  of  tri- 
angles as  radius  strike  a  small  arc,  2',  which  arc  is  to 
be  intersected  with  one  struck  from  T  of  pattern  and 
the  distance  H  2  of  plan  as  radius.  Proceed  in  this 
manner,  using  the  spaces  in  H  G  of  plan  for  the  dis- 
tances in  T  S  of  pattern,  and  the  lengths  of  lines  drawn 
from  M  to  points  2  to  5  in  diagram  of  triangles  for  the 
distances  across  the  pattern  from  Q  to  the  points  in 
T  S.  With  S  of  pattern  as  center,  and  G  F  of  plan  as 
radius,  describe  a  small  arc,  R,  which  intersect  with 
one  struck  from  Q  of  pattern  as  center,  and  M  C'  of  the 
triangles  or  B  C  of  the  elevation  as  radius,  thus  estab- 


lishing the  point  R  of  pattern.     Draw  R  S,  and  trace  :i 
line   through   the  points  from  S  to  T,  as  shown.      The 

Q 


_—  —  —                                  1 

1 

~"        — 

\\ 

1 

v\ 

1 

\  \ 

I 

\  \ 

1 

\   \ 

1 

\    \ 

1 

\     \          PATTERN 

/ 

\       \ 

1 

1 

\        \ 

1 

1 

\            \ 

\      \ 

\               \ 

1 

•\                  \ 

\                   \ 

1 

1 

V.                         / 

\                    \ 
\ 

v       / 

\ 

\.     / 

\              J 

\      jS" 

V                               I 

j                                              1         2 

T 

Fig.  535. — Pattern  of  Con-r. 

other  part  of  pattern,  as  Q  V  W  P,  can  be  described 
in  the  same  manner,  or  by  duplication. 


PROBLEM    166. 


Pattern  for  the  End  of  an  Oblong  Vessel  which  is  Semicircular  at  the  Top  and  Rectangular  at  the  Bottom. 


In  Fig.  536,  A  C  D  E  represents  the  side  eleva- 
tion of  the  article,  F  H  K  L  the  end  elevation,  and  M 
N  R  P  in  Fig.  537  the  plan.  By  inspection  of  these 
it  will  be  seen  that  the  portion  represented  upon  the 


SIDE. 


END. 


Fig.  536. — Side  and  End  Elerations  of  Oblong  Vessel  with 
Semicircular  End. 


end  elevation  by  G  L  K  is  simply  a  flat  triangular  sur- 
face, while  the  corners  of  the  vessel,  shown  by  B  C  D 


of  the  side  view  and  N  R  of  the  plan,  are  quarters  of 
the  envelope  of  an  inverted  scalene  cone. 

To  obtain  the  patterns  proceed  as  follows :  Divide 
one-half  of  the  end  of  the  plan  into  any  convenient 


Fig.  537. —Plan  of  Oblong  Vessel  with  Semicircular  End. 

number  of  equal  spaces,  all  as  shown  by  small  figures 
1,  2,  3,  4,  etc.,  in  N  R.      From  eacli  of  the  points  thus 


Patkrn   Problems. 


311 


determined  draw  linos  to  the  point  X,  the  apex  of  the 
cone  in  plan,  all  as  shown  in  the  engraving.  1'rocerd 
next  to  construct  the  diagram  of  triangles  shown  in 
Fig.  538,  of  which  the  lines  just  drawn  in  the  plan  are 
the  bases  and  15  I)  is  the  common  altitude.  Draw  A  B, 
in  length  equal  to  D  B  of  Fig.  536,  and  at  right  angles 
to  it  draw  B  C,  which  produce  indefinitely.  From  B 


54321, 


Fig.  5SS. — Diagram  of 
Triamjlfx. 


Fiij.  539.—  Pattern  of 
End  Pieces. 


along  B  C  set  off  spaces  equal  to  the  distances  from  X 
of  the  plan  to  the  several  points  in  the  boundary  line. 
That  is,  make  B  5  of  Fig.  538  equal  to  N  5  of  Fig. 
537,  and  B  4  equal  to  N  4,  and  so  on.  And  from 
each  of  the  points  in  B  C  draw  lines  to  A.  Then  the 
distances  from  A  to  the  various  points  in  B  C  will  be 


tin!  distances  from  the  apex  of  the  scalene  cone  to  the 
various  points  assumed  in  its  base,  and  will  be  the 
radii  of  the  arcs  shown  between  D  F  and  E  in  Fig. 
539.  For  convenience  erect  any  perpendicular,  as  A1 
Dof  Fig.  539,  upon  which  set  off  distances  equal  to  the 
length  of  the  lines  in  the  diagram  drawn  from  A,  or, 
in  other  words,  make  A1  1  equal  to  A  1  of  the  diagram, 
Fig.  538  ;  A1  2  equal  to  A  2  of  the  diagram,  and  so  on. 
From  A1  as  center,  with  radius  A1  D,  describe  the  arc 
D  E  indefinitely.  In  like  manner,  from  the  same 
center,  witli  radius  A1  2,  describe  a  corresponding  arc, 
and  proceed  in  this  way  with  each  of  the  other  points 
lying  in  the  line  A1  D. 

From  any  convenient  point  upon  arc  6,  as  F,  draw 
A1  F,  which  will  represent  the  side  of  the  pattern  cor- 
responding to  B  D  of  the  side  elevation.  With  the 
dividers  set  to  the  space  used  in  stepping  off  the  arc 
N  R  of  the  plan,  place  one  foot  at  the  point  F  of  the 
pattern  and  step  from  one  arc  to  the  next  until  all  the 
arcs  are  reached,  and  draw  A1  E.  Then  A'  F  E  will 
be  one  portion  of  the  required  pattern.  From  E  as 
center,  with  radius  E  A1,  describe  the  arc  A'  G  in- 
definitely. Make  the  chord  A1  G  equal  to  L  K  of  the 
end  elevation,  Fig.  536,  and  draw  E  G.  Then  A1  E  G 
will  be  the  pattern  of  that  portion  shown  by  L  G  K  of 
the  end  view  and  N  K  P  of  the  plan.  Duplicate  the 
part  A'  F  E,  as  shown  by  G  H  E,  thus  completing  the 
pattern  of  the  entire  end. 


PROBLEM  167. 

Pattern  of  an  Irregular  Flaring:  Article,  both  Top  and  Bottom  of  which  are  Round  and  Parallel,  but 
Placed  Eccentrically  in  Plan ;  Otherwise  the  Envelope  of  the  Frustum  of  a  Scalene  Cone. 


In  Fig.  540  is  shown  an  elevation  and  plan  of  the 
article,  in  which  E  F  G  II  is  the  plan  of  the  bottom 
and  E  J  K  L  that  of  the  top,  the  two  being  tangent  at 
the  point  E.  In  Fig.  541  the  elevation  and  a  portion 
of  the  plan  are  drawn  to  a  larger  scale  and  conven- 
iently located  for  describing  the  pattern. 

Since  the  top  and  the  base  of  the  article  are  both 
circular  and  are  parallel,  the  shape  of  which  the  pat- 
tern is  required  becomes  a  frustum  of  a  scalene  cone, 
and  lines  drawn  upon  its  surface  from  any  set  of  points 
assumed  in  the  circiimference  of  its  base  to  its  apex 
will  divide  the  circumference  of  the  top  into  similar 


and  proportionate  spaces.  Therefore,  the  first  step  is 
to  extend  the  lines  of  the  sides  B  A  and  C  D  until 
they  meet  at  M,  the  apex.  Next  divide  the  plan  of 
the  base,  one-half  of  which,  E  II  G,  only  is  shown, 
into  any  convenient  number  of  equal  spaces,  as  shown 
by  the  small  figures.  As  it  is  necessary  to  ascertain 
the  distance  from  each  of  these  points  to  the  apex  of 
the  cone  the  simplest  method  of  accomplishing  this  is 
as  follows :  From  E,  the  position  of  the  apex  in  plan, 
as  a  center,  with  E  0,  E  5,  E  4,  etc.,  as  radii,  describe 
arcs  cutting  E  G.  Carry  lines  vertically  from  each  of 
the  points  in  E  G,  cutting  the  base  line  A  D;  thence 


312 


The  New  Metal   Worker  Pattern  Book. 


carry  them  toward  the  apex  M,  cutting  the  line  of  the 
top  B  C,  all  as  shown. 

With  M  as  center  describe  arcs  from  each  of  the 
points  in  the  base  line  A  D,  and  extend  them  indefi- 
nitely in  the  direction  of  0.  In  the  same  manner 
draw  arcs  from  the  points  of  intersection  in  B  C,  as 
shown.  From  the  apex  M  draw  any  line  to  intersect 
the  arc  from  A  or  7  of  the  base  line,  as  M  N,  which 
will  form  one  side  of  the  pattern,  corresponding  to 
B  A  of  the  elevation.  Set  the  dividers  to  the  space 


drawn  from  BC;   then  a  line    traced   through  these 
points  of  intersection,  as  shown  by  E  P,  will  be  the  top 


Fig.  540. — Plan  and  Elevation  of  Frustum  of  a  Scalene  Cone. 


E  6,  used  in  dividing  the  plan  of  the  base,  ana  starting 
from  N  step  from  one  arc  to  the  next,  thus  laying  out 
the  stretchout  of  the  base  E  H  Gr,  and  at  the  same 
time  locating  each  point  at  its  proper  distance  from  the 
apex  M.  A  line  traced  through  these  points,  as  N  Q 
0,  will  be  the  bottom  of  one-half  of  the  pattern. 
From  the  points  in  the  line  N  Q  0  draw  lines  to  M, 
cutting  the  arcs  of  corresponding  number  previously 


E7 


Fig.  541. — Method  of  Obtaining  Pattern  of  a  Scalene  Cone. 

of  the  pattern,  and  P  R  N  Q  0  will  thus  be  one-half 
the  required  pattern. 


PROBLEM  168. 

Pattern  of  a  Flaring  Article,  the  Top  of  which  is  Round  and  the  Bottom  Oblong:  with  Semicircular 

Ends.— Two  Cases. 


First  Case. — In  Fig.  542  is  shown  the  elevation 
and  plan  of  the  article  drawn  in  proper  relation  to  each 
other,  as  shown  by  the  lines  of  projection.  In  this  case 


the  top  of  the  article  is  located  centrally  with  reference 
to  the  bottom,  as  shown  in  the  plan.  From  O,  the 
center  of  the  top,  erect  tne  perpendicular  0  o,  cutting 


Pattern  Problems. 


313 


the  line  of  the  top  in  elevation,  and  from  P  erect  the 
perpendicular  P/»,  cutting  the  line  of  the  base.  Since 
L  M  N  and  F  II  G  of  the  plan  are  semicircles  lying  in 
parallel  planes,  that  part  of  the  pattern  of  the  article 
shown  Lv  L  F II  G  N  M  must  be  one-half  the  envelope 
of  the  frustum  of  a  scalene  cone. 


Fig.  542.— Plan,  Elevation  and  Pattern  of  Flaring  Article  with  Kounil 
Top  and  Oblong  Base,  the  Top  being  Centrally  Located. 


To  ascertain  the  apex  of  the  cone,  prolong  the  side 
line  C  D  of  the  elevation  indefinitely  in  the  direction 
of  X.  Through  the  points  p  and  o  draw  p  o,  which 
produce  until  it  meets  C  D  prolonged  in  the  point  X. 
Then  X  is  the  apex  required.  From  X  drop  a  per- 
pendicular, cutting  the  center  line  of  the  plan  at  X1, 
thus  locating  the  apex  in  plan.  Divide  F  H  G  into 


any  convenient  number  of  equal  spaces,  as  shown  -by 
the  small  figures.  Should  lines  be  drawn  from  each  of 
the  points  thus  obtained  to  X1,  they  would  represent 
the  bases  of  a  set  of  right  angled  triangles,  of  which  Y 
X  is  the  common  altitude,  and  whose  hypothenuses 
will  give  correct  distances  from  the  apex  of  the  cone 
to  the  various  points  assumed  in  the  base. 

The  simplest  method  of  obtaining  these  hypothe- 
nuses is  as  follows  :  From  X'  as  center  draw  arcs  from 
each  of  the  points  in  F  H  G,  cutting  the  center  line 
K  II  of  the  plan.  From  each  point  in  P  H  erect  a  per- 
pendicular to  p  C,  as  shown.  From  the  points  thus 
obtained  in^>  C  carry  lines  toward  the  apex  X,  cutting 
o  D,  as  shown.  From  X  as  center  strike  arcs  from 
each  of  the  points  in  p  G  indefinitely.  Assume  any 
point,  as  G1,  upon  the  arc  struck  from  point  1  as  the  first 
point  in  the  pattern  of  the  base,  from  which  draw  a  line 
to  X.  Set  the  dividers  to  the  space  used  in  stepping 
off  the  plan,  and,  commencing  at  G1,  step  to  the  second 
arc,  and  from  that  point  to  the  third  arc,  and  so  on,  as 
shown  in  the  engraving.  A  line  traced  through  these 
points  will  be  the  boundary  of  a  lower  side  of  the  semi- 
circular end.  From  each  of  these  points  just  obtained 
draw  a  line  toward  the  center  X.  Place  one  foot  of  i 
the  dividers  at  X,  and,  bringing  the  pencil  point  suc- 
cessively to  the  points  in  o  D,  cut  radial  lines  of  cor- 
responding number  just  drawn.  A  line  traced  through 
these  points  of  intersection,  as  shown  by  N1  L1,  will 
form  the  upper  edge  of  the  pattern  of  the  end  piece. 
From  the  point  L1,  which  corresponds  to  L  of  the 
plan,  as  center,  with  L'  F'  as  radius,  describe  the  arc 
F'  E1,  and  from  F1  as  center,  with  radius  equal  to  F  B 
of  the  plan,  intersect  it  at  E1,  as  shown.  Draw  L1  R1. 
Then  L'  E1  F1  is  the  pattern  of  one  of  the  sides.  To 
L'  R1  add  a  duplicate  of  the  end  piece  already  ob- 
tained, all  as  shown  by  L1  E1  S'  N",  and  to  S'  N"  add 
a  duplicate  of  the  side  just  obtained,"  as  shown  by 
S1  N'  G*,  thus  completing  the  pattern. 

Second  Case. — This  case  differs  from  the  first  onlv 
in  the  fact  that  the  top  of  the  article,  being  located  near 
one  end,  is  drawn  concentric  with  the  semicircle  of  the 
near  end.  As  the  result  of  this  condition,  that  portion 
of  its  pattern  shown  by  S  E  R  L  K  N  in  the  plan,  Fig. 
543,  becomes  one-half  the  envelope  of  the  frustum  of 
a  right  cone,  the  method  of  developing  which  is  given 
in  Problem  123  of  the  previous  section  of  this  chapter. 

In  that  portion  of  the  article  shown  by  E  F  H  G 
S  N  the  conditions  are  exactly  the  same  as  in  the  first 
case.  In  Fig.  543  corresponding  parts  have  been  let- 
tered the  same  as  in  Fig.  5423  so  that  the  demonstra- 


.314 


Tlie  New  Metal    Worker  Pattern  Book. 


tion  given  above  is  equally  applicable  to  either  figure. 
In  the  final  make  up  of  the  various  parts  of  which  the 


a  duplicate  of  L'  F1  G'  N1,  all  as  clearly  shown   in   the 
engraving.     The  seam  in  this  case  has  been  located 


Rf 


I'-- - 


Fig.  54S.— Plan,  Elevation  and  Pattern  of  Flaring  Article  with  Round  Top  and  Oblong  Base,  the  Top  being  Located 

near  One  End. 


complete  pattern  is  composed,  the  part  R1  S1  N"  L1  is 
of  course  obtained  as  in  Problem  123  instead  of  being 


upon  the  line  N  S  of  the  plan  instead  of  upon  N  G  as 
before. 


PROBLEM    169. 

The  Patterns  of  a  Flaring  Tub  with  Tapering  Sides  and  Semicircular  Head,  the  Head  having:  More 

Flare  than  the  Sides. 


In  Fig.  544,  A  B  C  D  shows  a  side  elevation  of 
the  tub,  L  M  N  0  P  the  plan  at  the  top,  and  E  F  G 
II  K  the  plan  at  the  bottom,  an  inspection  of  which 
will  show  that  the  head,  as  shown  by  II  O  or  C  D,  has 
more  flare  than  the  sides,  whose  flare  is.  shown  by  A  J 
or  A  B,  the  flare  of  the  sides  and  foot  being  the  same. 
Inasmuch  as  the  article  is  tapering  in  plan,  the  conical 
part  of  the  pattern  will  include  a  little  more  than 


a  semicircle,  as  shown.  The  points  showing  the 
junction  between  the  straight  sides  and  the  conical 
part  are  to  be  determined  by  lines  drawn  from  the  cen- 
ters by  which  the  top  and  bottom  were  struck,  per- 
pendicular to  the  sides  of  the  article.  Therefore  lay 
off  in  the  plan  T  N  and  T  P,  drawn  from  the  center  T 
of  the  curved  part  of  the  plan  of  the  top  of  the  article, 
perpendicular  to  the  sides  M  N  and  L  P  respectively. 


Pattern  Problems. 


31.5 


And  in  like  manner  from  S,  the  center  by  which  the 
curved  part  of  the  bottom  of  the  article  is  struck, 
draw  S  <j  and  S  K. 

Since  the  top  and  bottom  of  the  tub  are  parallel, 
as  shown  by  the  side  elevation,  and  their  circles  are 
not  concentric  in  the  plan,  it  follows  that  the  part  P  0 


Fiij.   ~>44.—l1tin,  Elevation  and  J'attern  of  Flarituj  Tub  with.  'laperimj  Sides  and 

Semicircular  Head. 


N  G  II  K  is  part  of  the  envelope  of  a  scalene  cone. 
To  lind  the  apex  of  this  cone,  first  drop  lines  from  the 
points  T  and  S  vertically,  cutting  respectively  the  top 
and  bottom  lines  of  the  elevation,  as  shown  at  T1  and 
S1.  A  line  connecting  T1  and  S'  will  give  the  inclina- 
tion of  the  axis  of  the  cone  in  that  view,  which  con- 
tinue indefinitely  in  the  direction  of  R1  until  it  inter- 
sects the  side  C  D  continued,  as  shown  at  R'.  Then 


R1  is  the  apex  of  the  cone.  From  R'  draw  R1  R  verti- 
cally, cutting  the  center  line  of  the  plan  at  R.  Then 
R  shows  the  position  of  the  apex  of  the  cone  in  the 
plan.  As  the  pattern  of  the  curved  portion  consists 
of  two  symmetrical  halves  when  divided  by  the  center 
lino  of  the  plan,  divide  the  curve  N  0  into  any  con- 
venient number  of  equal  spaces,  as  shown  by 
the  small  figures.  Lines  drawn  from  each 
of  these  points  to  R  would  represent  the 
bases  of  a  series  of  right  angled  triangles 
whose  common  altitude  is  V  R1,  and  whose 
hypothenuses  when  drawn  will  represent  the 
correct  distances  from  the  apex  to  the  vari- 
ous points  assumed  in  the  base  of  the  cone. 
The  simplest  method,  however,  of  meas- 
uring these  bases  is  to  place  one  foot  of  the 
compasses  at  the  point  R,  and,  bringing  the 
pencil  point  successively  to  the  points  in  N 
O,  draw  arcs  cutting  the  center  line,  as  shown 
between  T  and  0.  Now  place  the  blade  of 
the  J-square  parallel  to  R  R1  and  drop  lines 
from  each  of  these  points,  cutting  the  line 
A  D  as  shown.  From  the  points  obtained 
upon  A  D  draw  lines  toward  the  apex  R', 
cutting  the  bottom  line  of  the  tub  B  C. 
These  lines  drawn  from  the  points  in  A  D 
to  R1  will  be  the  desired  hypothenuses  and 
may  be  used  in  connection  with  the  spaces 
of  the  plan  in  developing  the  envelope  of  the 
scalene  cone. 

Therefore  from  R1  as  center,  and  radii 
corresponding  to  the  distance  from  R1  to 
the  several  points  in  T1  D,  describe  a  set  of 
arcs  indefinitely,  as  shown.  Assume  any 
point  upon  the  arc  0,  as  N',  as  a  starting 
point,  from  which  draw  a  line  to  R'.  With 
the  dividers  set  to  the  space  used  in  divid- 
ing the  plan  N  O,  place  one  foot  at  the  point 
N1  and  swing  the  other  foot  around,  cutting 
the  arc  1.  Repeat  this  operation,  cutting 
the  arc  2,  and  so  continue  to  step  from  arc 
to  arc  until  all  the  arcs  have  been  reached, 
which  will  complete  the  outline  of  one-half  the  pattern. 
The  operation  of  stepping  from  arc  to  arc  can  be  con- 
tinued, stepping  back  from  arc  5  till  arc  0  is  reached 
at  P1,  thus  completing  the  top  line  of  the  pattern  of  the 
entire  curved  portion  of  the  tub.  From  each  of  the 
points  thus  obtained  draw  lines  toward  the  apex  R1,  as 
shown.  Place  one  foot  of  the  compasses  at  R',  and, 
bringing  the  pencil  point  successively  to  the  points  in 


316 


The  New  Metal    Worker  Pattern  Book. 


the  line  S'  C  previously  obtained,  cut  radial  lines  of 
corresponding  number  in  the  pattern,  as  shown  from 
G'  to  K1.  Lines  traced  through  the  several  points  in 
the  two  outlines,  as  shown  by  G'  H1  K1  and  N1  0'  P', 


will  complete  the  pattern  of  the  conical  part  of  the  tub. 
The  patterns  of  the  sides  and  foot  mav  be  obtained 
as  described  in  Problem  74  and  as  indicated  in  the 
upper  part  of  the  engraving. 


PROBLEM   170. 

The  Pattern  of  a  Flaring;  Article  whici  is  Rectangular  with  Rounded  Corners,  Having:  More 

Flare  at  the  Ends  than  at   the  Sides. 


In  Fig.  545,  A  B  C  D  E  F  represents  the  plan  at 
the  top  of  a  portion  of  the  flaring  article,  whose  general 
shape  is  rectangular  with  rounded  corners.  G  II  I  J 
K  L  represents  the  plan  of  the  bottom  of  the  same, 
showing  that  the  flare  at  the  ends,  represented  by  I  G 
or  J  D,  is  greater  than  that  of  the  sides,  represented  by 
A  G.  The  arc  E  D  of  the  plan  of  the  top  is  struck 
from  0  as  center,  while  the  arc  J  K  of  the  bottom  is 
struck  from  N.  Since  the  top  and  bottom  are  parallel, 
as  shown  by  P  Q  and  P  S  of  the  side  elevation,  the 
corner  J  D  E  K  is  a  portion  of  the  envelope  of  the 
frustum  of  a  scalene  cone. 

To  find  the  apex  of  the  cone,  drop  lines  from  0 
and  N  at  right  angles  to  P  Q,  cutting  respectively  the 
top  and  bottom  lines  in  the  side  view,  as  shown  at  T 
and  U,  and  draw  TU  and  continue  the  same  indefinitely 
in  the  direction  of  X.  Also  continue  Q  S  until  it  in- 
tersects T  U  at  the  point  X.  Then  X  will  be  the  apex 
of  the  cone.  From  X  erect  a  line  vertically,  cutting  the 
line  D  M  of  the  plan  at  M ;  then  M  will  show  the  posi- 
tion of  the  apex  of  the  cone  in  plan.  Divide  the  arc  E 
D  into  any  convenient  number  of  equal  spaces,  and 
from  the  points  thus  obtained  draw  arcs  from  M  as 
center,  cutting  0  D,  as  shown.  From  the  points  in  0 
D  drop  lines  vertically,  cutting  T  Q,  the  top  line  of  the 
side,  otherwise  the  base  of  the  cone.  From  the  points 
thus  obtained  in  T  Q  draw  lines  toward  the  apex,  cut- 
ting U  S. 

With  one  foot  of  the  compasses  set  at  X,  bring  the 
pencil  point  successively  to  the  points  in  T  Qand  draw 
arcs  indefinitely,  as  shown.  From  any  convenient 
point  upon  the  arc  0,  as  E',  draw  a  line  to  X,  forming 
one  side  of  the  pattern.  Take  between  the  points  of 
the  dividers  a  space  equal  to  that  used  in  stepping  off 
the  plan  of  cone  E  D,  and,  placing  one  foot  at  the  point 
E1,  swing  the  other  foot  around.  Cutting  the  arc  1,  from 
which  intersection  step  to  the  next  arc,  and  so  continue 
until  all  the  arcs  have  been  reached  at  D',  from  which 


point  draw  a  line  to  tin-  'ipc.\  X.  Likewise  from  each  of 
llie  points  between  E'  and  1)'  draw  line.-1  toward  the  apex 
indefinitely.  Finally,  with  one  foot  of  the  compasses  at 
X,  bring  the  pencil  point  to  each  of  the  points  in  LT  S 


H/i 


PLAN 


N 

"T 
I 

JK 


M 

0 

i 
j 

/ 

l.'l 

-C 


Fig.  545. — Plan,  Elevation    and   Pattern  uf  Rounded  Corner  fur  tin 
Article  whose  Sides  and  Ends  Flare  Unequally. 


and  draw  arcs,  cutting  radial  lines  of  corresponding 
number  in  the  pattern.  Lines  traced  through  the 
points  between  the  points  K'  and  J'  and  E'  and  D'will 
complete  the  pattern  of  the  curved  corner.  Tin'  pat- 
tern for  the  plain  sides  can  easily  be  obtained  after  the 
manner  described  in  Problem  74  and  added  to  that  of 
the  corner  as  may  be  found  practicable. 


Pattern   Problems. 


317 


PROBLEM    171. 

t 

The  Pattern  of  a  Flaring  Article  which  Corresponds  to  the  Frustum  of  a  Cone  whose  Base  is  a  True 

Ellipse. 


In  Fig.  540,  let  G  H  F  E  be  the  elevation  of  one 
side  of  the  article,  L  M  U  R  the  elevation  of  an  end, 
K'  K'  F1  U'  the  plan  of  the  article  at  the  base,  and  T  V 
S  1'  the  plan  at  the  top.  Produce  E  Cr  and  F  II  of  the 
side  elevation  until  they  meet  in  the  point  I,  the  apex 
of  the  cone. 

Divide  one-quarter  of  the  plan  E1  R'  into  any  con- 


Fig.  54B. — Plan,  Side  and  End  Elevations  of  the  Frustum  of  an 
Elliptical  Cone. 


venient  number  of  equal  parts,  as  indicated  by  the 
small  figures.  From  the  points  thus  determined  draw 
lines  to  the  center  C.  These  lines  will  form  the  bases 
of  a  series  of  right  angled  triangles  whose  common  alti- 
tude is  the  hight  of  the  cone,  and  whose  hypothenuses 
when  drawn  will  give  the  true  distances  from  the  apex 
p  to  tiie  several  points  assumed  in  the  base  of  the  cone. 
Therefore  at  any  convenient  place  draw  the  straight 
line  D  A  of  Fig.  54-7,  in  length  equal  to  I  E';.  Make 
D  B  equal  to  1  G'.  From  A  and  B  of  Fig.  547  draw 
perpendiculars  to  D  A  indefinitely,  as  shown  by  A  O 


and  B  N.  Take  the  distances  C  5,  C  4,  C  3,  etc.,  of 
the  plan  and  set  off  corresponding  distances  from  A 
on  A  0,  as  shown  by  A  5,  A  4,  A  3,  etc.  From  these 
points  in  A  O  draw  lines  to  D,  cutting  B  N.  These 
lines  are  also  shown  in  the  elevation,  but  are  not 
necessary  in  that  view  in  obtaining  the  pattern.  From 
])  as  center  describe  arcs  whose  radii  are  equal  to  the 
lengths  of  the  several  lines  just  drawn  from  D  to  the 
points  in  A  0. 

From  any  convenient  point  in  the  first  arc  draw  a 


Fig.  547.— Diagram  of  Triangles  and  Pattern  of  Frustum 
in  Fig.  5j6. 


straight  line  to  D,  as  shown  by  W  D.  This  will  form 
one  side  of  the  pattern.  From  W,  as  a  starting  point, 
lay  off  the  stretchout  of  the  plan  E1  R'  F',  etc.,  using 
the  same  length  of  spaces  as  employed  in  dividing  it, 
stepping  from  one  arc  to  the  next  each  time,  as  shown. 
A  line  traced  through  these  points  will  be  the  outline 
of  the  base  of  the  pattern,  one-half  of  the  entire  en- 
velope being  shown  in  the  pattern  from  W  to  Z. 

From  the  points  in  W  Z  draw  lines  to  D,  which 
intersect  by  arcs  drawn  with  D  as  center  and  starting 
from  points  of  corresponding  number  in  B  N.  A 
line  traced  through  the  points  of  intersection  will 
form  the  upper  line  of  the  pattern,  as  shown.  Then 
\V  X  Y  /  will  constitute  the  pattern  of  one-half  of 
the  envelope,  to  which  add  a  duplicate  of  itself  for 
the  complete  pattern. 


318 


T 'lie  Xcw  Jtfdal    Worker   Pattern  Hook. 
PROBLEM  172. 


The  Patterns  for  a  Hip   Bath. 


In  Fig.  548,  let  II  A  L  N  0  bo  the  elevation  of 
the  bath,  of  whieh  D1  G  E1  IV  is  ;i  plan  on  the  line  1) 
E.  Let  the  half  section  A'  M  C'  BJ  represent  the  flare 
whieh  the  bath  is  required  to  have  through  its  sides 
on  a  line  indicated  by  A  B  in  elevation.  By  inspec- 
tion of  the  elevation  it  will  be  seen  that  three  patterns 
are  required,  which,  for  the  sake  of  convenience,  have 
been  numbered  in  the  various  representations  1,  2 
and  3. 

Since  the   plan   of  piece  No.  1   on  the  line  1)  B, 


M 


VF- 


Fig.  548.— Plan,  Elevation  and  Section  of  a  Hip  Hath. 


which  is  at  right  angles  to  its  axis  A  F,  is  a  semicircle, 
as  shown  by  G  D1  B',  and  since  its  Hare  at  the  side,  as 
shown  by  C'  B"  A",  is  the  same  as  at  B  D  H,  its  pat- 
tern will  be  a  portion  of  the  envelope  of  a  right  cone. 
Patterns  of  this  class  have  been  treated  in  the  previous 
section  of  this  chapter — to  which  the  reader  is  re- 
ferred— where,  in  Problem  143,  an  exactly  similar 
subject  has  been  treated.  The  operation  of  obtaining 


this  pattern  is  fully  shown  in  Fig.  549,  and,  therefore, 
need  not  be  here  described. 

Piece  Xo.  I',  as  shown  in  Fig.  548,  is  so  drawn  as 
to  form  one-half  of  the  frustum  of  an  elliptical  cone. 
As  its  section  at  A  B  (shown  at  the  right)  must  neces- 
sarily be  the  same  as  that  of  piece  No.  1,  against 
which  it  fits,  the  point  F  is  assumed  as  the  apex  of  the 
elliptical  cone,  and  consequently  the  flare  at  the  foot, 
E  L,  is  determined  by  a  continuation  of  the  line  drawn 
from  the  apex  through  E.  Should  it  be  decided  to 
have  more  flare  at  the  foot  than  that  shown  by  E  L, 
the  point  L  may  be  located  at  pleasure,  and  the  plan 
of  the  top,  K  L1  A1,  be  drawn  arbitrarilv ;  after  which 
its  pattern  may  be  developed  by  means  of  the  alter- 
nating triangles  alluded  to  in  the  introduction  of  this 


Fiij.  549.— Pattern  of  Piece  No.  1  of  Hip  Bath. 

section  (page  306),  examples  of  which  will  be  found 
further  on  in  this  section. 

The  plan  G  E'  B',  from  which  the  dimensions  of 
the  pattern  are  to  be  determined,  may  be  a  true  ellipse, 
or  may  be  composed  of  arcs  of  circles,  as  shown,  ac- 
cording to  convenience.  Divide  one-half  the  plan  G 
K1  into  any  convenient  number  of  equal  spaces,  as 
shown  by  the  small  figures,  and  from  each  point  thus 
obtained  draw  lines  to  the  center  C.  To  avoid  con- 
fusion of  lines  a  separate  diagram  of  triangles  is  con- 
structed in  Fig.  .550,  in  which  M  0  is  the  hight  of  tho 


Pattern  Pmblems, 


310 


tiii)  or  frustum  and  M  F  the  hight  of  the  cone.     Draw 

M   L  and  ('  K,  each  at  right  angles  to  M  F.    Upon  ('  K 
set  off   from  C  the  lengths  of  the  several  lines  C  1, 


1584 s    >L 


Fiy.  550.— Pattern  of  Piece  No.  2  of  Hip  Bath. 

('  -1,  etc-.,  of  the  plan.      Through  each  of  the  points  in 
C  E  draw  lines  from  F,  cutting  the   line  M  L.      From 


Fiy.  551.— Diagram  <>/  Radii  for  Pattern,  of  Foot. 

V   as  center   draw  ares    indefinitely  from  each  of  the 
points   in    C  E  and    also    from    the    points   in   M   L. 


Through  any  point  upon  arc  1  of  the  lower  set,  as  G, 
draw  a  line  from  F  and  extend  it  till  it  cuts  arc  1  of 
the  upper  set  at  K;  then  K  G  will  be  one  side  of  the 
pattern.  With  the  dividers  set  to  the  space  used  in 
dividing  the  plan  G  E1,  place  one  foot  at  the  point  G 
of  the  pattern  and  step  to  arc  2,  and  so  continue  step- 
ping from  one  arc  to  the  next  till  all  are  reached,  as 
;it  E,  and  repeat  the  operation  in  the  reverse  order, 
finally  reaching  B  and  completing  the  lower  line  of 
the  pattern.  From  each  of  the  points  in  G  E  B  draw 
lines  radially  from  F,  cutting  arcs  of  corresponding 


V  V- H- 


Fig.  552.— Pattern  of  Foot  of  Hip  Bath. 


number  drawn  from  M  L.  Lines  traced  through  these 
points  of  intersection  will  complete  the  upper  line  of 
the  pattern.  Then  G  E  B  A  L  K  will  be  the  required 
pattern  of  piece  No.  2. 

As  the  plan  D1  G  E1  B'  has  been  drawn  entirely 
from  centers  (C,  P,  S  and  P'),  the  pattern  of  piece 
No.  3  is  exactly  similar  to  that  described  in  Problem 
134  of  the  previous  section  of  this  chapter,  to  which 
the  reader  is  referred.  In  Fig.  551  is  shown  a  diagram 
for  obtaining  the  radii  taken  from  dimensions  given  in 
Fig.  548,  while  Fig.  552  shows  the  pattern  described 
by  means  of  the  radii  given  in  Fig.  551. 


320 


Tlie  New  Metal   Worker  Pattern  Book. 


PROBLEM   173. 


The  Patterns  for  a  Soapmaker's  Float. 


In  Fig.  553  is  shown  a  perspective  view  of  a  soap- 
maker's  float.  In  general  characteristics  it  is  very 
similar  to  piece  No.  2  of  the  hip  bath  treated  in  the 
preceding  problem.  It  also  resembles  the  bathtub  in 
that  its  bottom  is  bulged  or  raised  with  the  hammer, 
and  is  therefore  not  included  in  the  field  of  accurate 
pattern  cutting.  The  sides  are  to  be  considered  as 
parts  of  two  cones  having  elliptical  bases,  the  short 
diameters  of  which  are  alike,  but  the  long  diameters 
of  which  vary. 

In  Fig.  554  is  shown  a  plan  and  an  inverted  eleva- 
tion of  the  flaring  sides,  showing  in  dotted  lines  the 
completed  cones  of  which  the  sides  form  a  part.  Thus 
L  D1  M  represents  one-half  the  base  of  an  elliptical 
cone  of  which  L  M  is  the  short  diameter  and  D1  K1 
one-half  the  long  diameter.  As  all  sections  of  a  cir- 


fig.  553.— Perspective  View  of  Soapmaker's  Float. 

cular  cone  taken  parallel  to  its  base  are  perfect  circles, 
so  all  sections  of  an  elliptical  cone  parallel  to  its  base 
must  be  ellipses  of  like  proportions  with  the  base. 
Therefore  the  plan  of  the  upper  base  of  the  frustum 
A'  P  must  be  so  drawn  that  a  straight  line  from  A1  to 
P  will  be  parallel  to  a  straight  line  joining  D1  and  L. 

For  the  pattern  of  the  portion  shown  by  D  A  E  F 
of  the  elevation,  first  produce  the  line  F  E  of  the  ele- 
vation in  the  direction  of  K  indefinitely.  In  like  man- 
ner produce  D  A  of  the  elevation  until  it  reaches  F  E 
produced  in  the  point  K.  Then  D  K  F  may  be  re- 
garded as  the  elevation  of  a  half  cone,  of  which  that  part 
of  the  vessel  is  a  portion,  and  K  F  its  perpendicular 
hight.  Next,  divide  one-half  the  plan  L  D1  M  into 
any  number  of  equal  parts,  as  shown  by  the  small 
figures  1,  2,  3,  etc.  Construct  the  diagram  of  trian- 


gles, shown  in  Fig.  555,  by  drawing  the  line  D  K1  of 
indefinite  length,  and  the  line  K'  K  at  right  angles  to 
it,  making  K1  K  in  length  equal  to  F  K,  Fig.  554. 
Establish  the  point  K2  by  making  the  distance  K1  K'J 
equal  to  E  F  of  Fig.  554.  Draw  K2  A  parallel  to  K'  1). 
From  each  of  the  points  2,  3,  4,  etc.,  of  the  plan  draw 
lines  to  the  center  K1,  and  set  off  distances  equal  to 
these  lines  upon  the  line  K1  D  of  Fig.  555,  measuring 
from  K'  toward  D.  From  eacli  of  the  points  thus  ob- 
tained draw  lines  to  the  point  K,  cutting  A  K'.  With 
one  foot  of  the  compasses  in  the  point  K,  and  the  other 


Fig.  554. — Plan  and  Inverted  Elevation  of  Soapmaker's  Float. 

brought  successively  to  the  points  1,  2,  3,  etc.,  in  the 
line  D  K'  and  also  to  the  points  in  the  line  A  K2,  de- 
scribe arcs  indefinitely. 

Take  in  the  dividers  a  space  equal  to  the  divisions 
in  D1  L  of  Fig.  554,  and,  commencing  at  the  point  a 
in  arc  7  (Fig.  555),  step  to  arc  6  and  thence  to  arc  5, 
and  thus  continue  stepping  from  one  arc  to  the  next 
until  the  entire  stretchout  of  the  half  plan  has  been 
laid  off,  as  shown  in  Fig.  555.  From  each  of  the 
points  thus  obtained  in  a  d  draw  lines  to  K,  cutting 
arcs  of  corresponding  number  drawn,  from  A  K2.  Then 


Pattern  Problems. 


321 


a  line  traced  through  the  several  points  of  intersection 
thus  obtained,  as  shown  by  l>  <•  and  a  d,  will  be  the 
boundary  lines  of  tho  pattern. 

The  pattern  for  the  other  end  of  the  article  is  to 
be,  in  the  main,  developed  in  the  same  manner  as  above 
described.  One  additional  condition,  however,  exists 
in  connection  with  this  piece,  vi/.  :  To  determine  the 
dimensions  of  the  cone  of  which  this  piece  (K  15  C  F, 
Fig.  ;">54)  is  a  part,  since  the  flare  at  B  C  is  much 
greater  than  A  D,  while  1,1 10  flan;  of  both  pieces  at  the 
side  is  the  same  as  shown  by  P  L  of  the  plan.  The 
quarter  ellipse  P  B'  of  the  plan  being  given,  and  also 
the  point  C1,  it  becomes  necessary  to  draw  from  C1  a 
quarter  ellipse  which  shall  be  of  like  proportions  with 


Q123J  Mi? 


Fig.  555. — Diagram  of  Triangles  and  Pattern  of  Piece 
L  D>  M  of  FiAj.  554. 


P  B1,  which,  as  remarked  above,  is  a  necessary  condi- 
tion, both  being  horizontal  sections  of  the  same  cone. 
To  do  this  proceed  as  follows  :  Connect  the  points  P 
and  B1  by  means  of  a  straight  line.  From  the  point 
C'  draw  a  line  parallel  to  P  B1,  and  produce  it  until  it 
cuts  the  line  L  G',  which  is  a  straight  line  drawn  at 
right  angles  to  L  M.  Then  G1  becomes  a  point  in  the 
lower  base  of  the  cone  corresponding  to  the  point  P  in 
the  upper  base.  Draw  the  line  G'  P,  and  continue  it 
until  it  intersects  the  long  diameter  in  IT.  Drop  the 
point  G1  vertical  from  the  plan  on  to  the  base  line  D  C 
of  the  elevation,  as  indicated  by  the  point  G.  Draw  a 
line  through  the  points  G  and  E,  which  produce  in- 
definitely in  the  direction  of  H.  In  like  manner  pro- 


duce the  side  C  B  of  the  cone  until  it  intersects  G  E 
produced  in  tho  point  II.  Then  it  will  be  found  that 
the  point  II  of  the  elevation  and  the  point  H1  of  the 
plan  coincide,  as  indicated  by  the  line  H  H1.  H  G  of 
the  elevation  thus  represents  the  axis  and  H  0  one 
side  of  the  cone  of  which  the  piece  E  B  C  G  is  a  part, 
from  which  it  will  be  seen  that  this  cone  is  at  once 
elliptical  and  scalene.  .  The  operation  of  developing 
the  pattern  from  this  stage  forward  is  the  same  as  in 
the  previous  case,  all  as  clearly  shown  in  Fig.  556,  save 
only  in  the  addition  of  the  triangular  piece  indicated 
by  G  E  F  of  the  elevation.  After  completing  the  other 


[8   ?Q    ti    6    4 


Fig.  556. — Diagram  of  Triangles  and  Pattern  of  Piece 
L  &  M  of  Fig.  554. 


portions  of  the  pattern,  this  triangular  piece  is  added 
as  follows :  The  distance  H'  L  in  Fig.  556  is  to  be  set 
off  on  the  line  H'  C  in  the  same  manner  as  the  distances 
to  the  other  points — i.e.,  H1  L  is  equal  to  H1  L  of  Fig. 
554.  Then  L  is  to  be  treated  in  the  same  manner  as 
the  other  points,  an  arc  being  struck  from  it,  as  indi- 
cated in  the  engraving,  by  which  to  determine  the 
corresponding  point  L"  in  the  outline  of  the  pattern. 
L"  G3  is  made  equal  to  L  G1  of  the  plan,  Fig.  554. 
From  L'  draw  a  line  to  E.  Then  E  L"  G2  will  be  the 
pattern  of  the  triangular  piece  indicated  in  Fig.  554  by 
E  F  G.  It  is  to  be  added  upon  the  opposite  end  of 
the  pattern  in  like  manner,  as  indicated  by  E'  G1  L1. 


322 


Tlie  New  Metal   Worker  Pattern  Book. 


PROBLEM    174. 
The  Envelope  of  a  Frustum  of  an  Elliptical  Cone  Having  an  Irregular  Base. 


The  form  E  P  K  L  J  shown  in  Pig.  557,  the 
lower  line  of  which  is  an  irregular  section  through  an 
elliptical  cone,  is  introduced  here,  not  as  representing 
any  particular  article  or  class  of  forms,  but  because  it 
embodies  a  principle  somewhat  different  from  other 
sections  of  cones  previously  given,  which  may  be  use- 
ful to  the  pattern  cutter. 

B  A  C  is  the  side  elevation  of  a  cone  having 
an  elliptical  base,  one-half  of 
which  is  shown  by  B1  H  C1. 
Divide  one  quarter  of  the  plan, 
as  H  C',  into  any  convenient 
number  of  equal  parts,  as 
shown  by  the  small  figures. 
From  each  of  the  points  of 
division  draw  lines  to  the  cen- 
ter D",  and  also  erect  lines 
cutting  the  base  of  the  cone 
B  C,  from  which  carry  them 
toward  the  apex,  cutting  the 
lines'  E  F  and  J  L  K.  The 
lirst  operation  will  be  that  of 
obtaining  the  envelope  of  the 
complete  cone  in  the  same 
manner  as  described  in  pre- 
vious problems. 

Construct  a  diagram  of 
triangles,  as  shown  at  the 
right,  in  which  A1  D1  is  equal 
in  hight  to  A  D,  and  at  right 
angles  to  G1  W  and  D1  V,  ex- 
tensions respectively  of  E  F 
and  B  C.  Upon  D'  V,  measur- 
ing from  D1,  set  off  the  dis- 
tances from  D"  to  the  several  points  in  H  C1,  as 
shown.  From  each  of  the  points  thus  obtained  draw 
lines  toward  A1,  cutting  G1  W.  Also  from  each 
of  these  points,  with  A'  as  center,  describe  arcs  indefi- 
nitely. Take  between  the  points  of  the  dividers  a 
space  equal  to  that  used  in  dividing  the  plan  H  C', 
and  placing  one  foot  upon  the  arc  drawn  from  point  1 
in  the  line  D1  V,  step  to  arc  2,  thence  to  arc  3  and  so 
continue  till  one  quarter  of  the  stretchout  is  completed 
at  7,  and,  if  desirable,  continue  the  operation,  taking 
the  arcs  in  reverse  order,  thus  completing  the  outline 
of  one-half  the  envelope  of  the  cone,  as  shown  in  the 


engraving.  From  each  of  the  points  in  this  outline  or 
stretchout  draw  measuring  lines  toward  the  center  A'. 
Place  one  point  of  the  compasses  at  point  A1,  and, 
bringing  the  pencil  point  successively  to  the  several 
points  of  intersection  on  the  line  G1  W,  cut  measuring 
lines  of  corresponding  number,  as  shown  from  G1  to 
Ga.  Place  the  T-square  parallel  to  the  base  B  C,  and, 
bringing  it  successively  to  the  several  points  of  inter- 


'  123  4  6  87 

DIAGRAM  OF 
TRIANGLES 

-7 


Fig.  557.— Patterns  for  the  Frustum  of  an  Elliptical  Cone  Having  an  Irregular  Base. 


section  previously  obtained  in  the  curved  line  L  K, 
cut  lines  of  corresponding  number  drawn  from  the 
points  in  D1  V  to  A1,  as  shown  from  X  to  Y.  Finally, 
with  one  foot  of  the  compasses  at  A1,  bring  the  pencil 
point  to  each  of  the  points  of  intersection  last  obtained 
and  cut  corresponding  measuring  lines  in  the  pattern. 
Then  lines  traced  through  the  points  of  intersection,  as 
shown  from  L1  to  L3  and  from  G'  to  G',  will  complete 
the  pattern  of  one-half  the  frustum  E  F  K  L  J. 

Should  it  be  desirable  to  cut  a  pattern  to  fill  the 
end  J  L  K  of  the  frustum,  as  for  a  bottom  in  the 
same,  it  will  first  be  necessary  to  obtain  a 'correct  plan 


Pattern  Problems. 


323 


of  the  line  J  K  L.  To  accomplish  this,  set  off  on  the 
lines  D"  7,  D''  6,  etc.,  of  the  plan  the  lengths  of  the 
several  lines  of  corresponding  number  drawn  from  the 
lino  G1  D1  to  the  intersections  between  X  and  Y,  thus 
obtaining  the  desired  line  L3  K2.  Extend  the  center 
line  B'  C1  of  the  plan,  as  shown  at  the  right,  upon 
which  lay  off  a  stretchout  of  the  line  L  K,  taking  each 
of  the  spaces  separately  as  they  occur,  all  as  shown  by 


Z  K3,  through  which  draw  measuring  lines  at  right 
angles.  Place  the  T-square  parallel  to  B'  C',  and,  bring- 
ing it  to  the  several  points  in  the  line  L3  K',  cut  corre- 
sponding measuring  lines.  Then  a  line  traced  through 
the  points  of  intersection,  as  shown  by  L*  K3,  will  be 
the  pattern  of  one-quarter  of  the  desired  piece,  which 
may  be  duplicated  as  necessary  for  a  half  or  for  the 
entire  pattern  in  one  piece. 


PROBLEM    175. 

The  Patterns  of  the  Frustum  of  a  Scalene  Cone  Intersected  Obliquely  by  a  Cylinder,  their  Axes  Not 

Lying:  in  the  Same  Plane. 


In   Fig.    558,  let  A  B  C  D  represent  the  frustum 
of  an  oblique  cone,  and  T  S  K  V  U  the  cylinder  that 


A  D  and  B  C   are  the  outlines  of  the  slanting  sides. 
'  In  Fig.  559  E  F  G  H  shows  the  plan  of  the  frustum 


SECTIONS 


Fig.  558.— Front  Elevation  of  the  Frustum  of  a  Scalene  Cone 
Intersected  Obliquely  by  a  Cylinder. 

joins  the  same  at  the  angle  indicated.      The  view  here 
given  of  the  frustum  is  that  of  its  vertical  side,  so  that 


Fig.  559. — Elevation,  Plan  and  Sections  of  the  Frustum  of  a 
Scalene  Cone  Intersected  Obliquely  by  a  Cylinder. 

at  its  base  and  K  I  J  G  the   plan  of  the  top,    from 
which  the  side  elevation  is  projected  at  the  left,  D  C 


324 


27;e  Xew  Metal    Worker  Pattern  Book. 


being  the  base  and  A  D  the  vertical  side.  The  inter- 
secting cylinder  is  indicated  by  F  L  M  G,  and  its  pro- 
file by  N  0  P  Q.  The  diameter  of  cylinder  is  the 
same  as  that  of  top  of  frustum. 

Divide  the  profile  N  O  P  Q  into  any  convenient 
number  of  equal  parts,  and  from  the  points  thus  ob- 
tained carry  lines  parallel  with  G  M,  cutting  I  J  G 
and  F  G  of  plan,  as  shown.  As  the  points  in  the 
profile  of  the  cylinder  lie  in  four  vertical  planes,  indi- 


shown.  With  I"  and  d"  as  centers,  strike  the  arcs 
G  b'  ;ui(l  (!  d\  thus  forming  sections  of  the  cone  in 
plan  corresponding  with  a  l>  and  c  d  of.  elevation.  The, 
four  vertical  sections  above  referred  to  are  shown 
below  the  plan  by  I'  F',  e  f,  J'  </'  and  /*'  i'.  To  avoid 
a  confusion  of  lines,  the  method  of  obtaining  the  shapes 
is  shown  separately  in  Figs.  560  to  563,  in  which  the 
reference  letters  are  the  same  as  in  Fig.  559. 

To  obtain  the   shape  of    section  on  line  I  F 


111 


T 

Fig.  560.  Fig.  561. 

Sections  of  the  Frustum  of  Scalene  Cone  Corresponding  to  Divisions  in  Profile  of  Cylinder  in  Fig.  559. 


cated  by  the  lines  7,  8  6,  1  5,  and  2  4,  it  will  be  nec- 
essary, before  their  intersection  can  be  obtained,  to 
construct  four  vertical  sections  through  the  cone  upon 
the  lines  I  F,  e  /,  J  g,  and  h  i.  The  point  3  requires 
no  section;  it  being  flush  with  the  vertical  side  of  the 
cone,  must  intersect  somewhere  on  the  line  A  D.  To 
obtain  the  desired  sections  divide  A  D  of  elevation 
into  any  convenient  number  of  equal  parts,  and  from 
the  points  thus  obtained  erect  lines  parallel  to  the  base 
D  C,  cutting  B  C.  From  the  points  in  B  C  carry  lines 
parallel  with  A  G,  cutting  the  center  line  E  G,  as 


Fig.  560,  extend  E  G,  as  indicated  by  I'  D',  which 
make  equal  to  A  D  of  the  elevation,  with  its  points  of 
division  a  and  c.  From  the  points  in  I'  D'  erect  the 
perpendiculars  a  b,  c  d  and  D'  F'.  With  the  T  square 
placed  parallel  with  I'  D','  drop  lines  from  the  points 
in  I  F,  cutting  similar  lines  drawn  from  I'  D'.  A  line 
traced  through  the  points  of  intersection,  as  shown  by 
I'  F',  will  give  the  required  shape.  The  sections 
shown  in  Figs.  561,  562  and  563  are.  obtained  in  a 
similar  manner. 

Having  obtained  these  sections  of  the  cone  by  the 


Pattern  Problems. 


325 


above  method,  arrange  them  as  shown  below  the  plan 
in  Fig.  55!*.  An  inspection  of  the  plan  and  profile 
will  show  that  a  line  drawn  from  O  of  profile  will  cut 
section  I  F,  line  8  6  of  profile  will  cut  section  ef,  line 
N  P  of  profile  will  cut  section  J  g,  line  2  4  will  cut 
section  h  i,  and  a  line  from  Q  of  profile  will  cut  the 
vertical  side  represented  by  (r. 

In  connection  with  the  sections  in  Fig.  559  draw 
an  elevation  of  cylinder,  as  shown  by  S  R  V  U,  oppo- 
site the  end  of  which  draw  a  profile,  as  indicated  by 


PATTERN 


s" 

Fig.  564. — Method  of  Obtaining  Pattern,  of  Cylinder  Shown  in  Figs.  558  and  559. 


N'  Q'  P'  O',  commencing  the  divisions  at  the  point 
K  '.  From  the  several  points  in  the  profile  N'  0'  P'  Q' 
carry  lines  parallel  with  U  V  against  the  several  pro- 
files I'  F',  e'  f,  J'  g'  and  h'  i'  as  described  above  and 
as  indicated  by  the  small  figures  1  to  8.  A  line  traced 
through  these  points  of  intersection  will  give  the  rniter 
line.  A  duplicate  of  this  part  of  Fig.  559  is  presented 
in  Fig.  564  for  the  purpose  of  avoiding  a  confusion  of 
lines.  The  miter  line  drawn  through  the  intersecting 
lines  is  indicated  by  S  T  U. 

Having   now   the   profile  of   the   cylinder  and  the 


miter  line,  all  as  shown,  the  panem  of  the  cylinder  is 
obtained  in  accordance  with  the  principles  given  in 
numerous  examples  in  Section  1  of  this  Chapter,  and 
as  clearly  shown  in  Fig.  564. 

The  method  of  obtaining  the  envelope  of  the 
frustum,  and  the  opening  in  the  side  of  the  same  to  fit 
against  the  end  of  the  cylinder  just  obtained,  is  shown 
in  Fig.  565.  The  simple  envelope  of  the  frustum  is 
obtained  exactly  as  described  in  Problem  167,  as  will 
be  seen  by  a  comparison  of  Figs.  565  and  541.  To 
obtain  the  opening  in  its  side,  however,  involves  an 
operation  similar  to  that  given  in  the  problem  immedi- 
ately preceding.  A  B  C  D  of  Fig.  565  represents  a 
side  elevation  of  the  frustum,  as  shown  by  the  same 
letters  in  Fig.  559,  and  the  vertical  lines  drawn 
through  the  same,  designated  by  the  small  figures  at 
the  bottom,  are  the  lines  of  the  vertical  sections  ob- 
tained in  Figs.  560-563,  and  correspond  in  numbers 
to  the  divisions  in  the  profile  in  Fig.  559.  To  obtain 
the  elevation  of  the  opening,  set  off  on  each  of  these 
section  lines  the  hights  of  the  points  of  intersection 
occurring  on  corresponding  sections  as  they  appear  in 
Fig.  564.  Thus  upon  line  designated  at  the  bottom 
by  2  4,  set  off  the  vertical  hights  of  points  2  and  4  on 
section  h'  i',  Fig.  564,  which  section  corresponds  to  line 
2  4  of  the  profile,  as  shown  in  Fig.  559.  In  the  same 
manner  set  off  on  line  1  5  the  verti- 
cal hights  of  the  points  1  and  5  om 
section  J'  g' '.  Obtain  also  points  8 
and  6  from  section  e'  f  and  point  7 
from  section  I'  F',  all  as  shown  by 
the  small  figures.  A  line  traced 
through  these  points  will  give  the 
elevation  of  the  opening.  The  out- 
line of  the  opening  has  been  shown 
in  the  plan,  but  its  development  is 
not  necessary  to  the  subsequent 
work  of  obtaining  the  pattern. 

Divide  the  plan  of  the  base  of 

the  frustum  G  F  E  into  any  convenient  number  of  equal 
spaces,  as  indicated  by  the  small  letters.  As  an  accu- 
rate elevation  of  the  opening  has  now  been  obtained,  this 
operation  can  be  conducted  without  reference  to  any  of 
the  points  previously  used  in  obtaining  the  line  of  the 
opening.  Therefore  letters  have  been  used  in  the  divis- 
ions of  G  F  E  instead  of  figures  so  that  no  confusion 
may  arise.  From  each  of  these  points  of  division  draw 
lines  to  G,  which  represents  the  plan  of  the  apex  of  the 
cone.  Also  from  so  many  of  these  points  from  which 
lines  will  cut  the  line  of  the  opening,  as  a  to  /,  erect 


326 


Tfie  New  Metal   Worker  Pattern  Book. 


lines  vertically,  cutting  the  base  line  D  C,  as  shown  by 
corresponding  letters.  From  these  points  draw  lines 
toward  the  apex  of  the  cone  X,  cutting  the  line  of  the 
opening  in  the  elevation,  as  shown  but  not  lettered. 

Proceed  now  to  construct  the  diagram  of  triangles 
shown  at  the  right,  in  which  X'  D'  is  equal  to  and 
parallel  with  X  D,  and  in  which  A1  B1  and  D1  C'  are 
drawn  in  continuation  of  A  B  and  D  C,  as  shown. 


dividers  at  DJ,  step  to  arc  b,  thence  10  arc  c,  etc..  till 
arc  i  is  reached  at  G'.  A  line  traced  through  these 
points  will  give  the  lower  outline  of  the  half  of  the 
envelope  of  the  frustum  which  is  pierced  by  the  cylin- 
der. From  each  of  these  points  also  draw  lines  toward 
X1,  which  intersect  by  arcs  of  corresponding  number 
drawn  with  X1  as  center  from  the  line  A'  B1,  thus  ob- 
taining the  upper  line  of  the  envelope. 


c          a         c 
DIAGRAM  OF 
TRIANGLES 


Fig.  565. — Method  of  Obtaining  Opening  in  Side  of  Cone  to  Fit  End  of  Cylinder. 


Upon  D1  C1,  measuring  from  D',  set  off  the  several 
lengths  G  J,  G  c,  etc.,  of  the  plan,  as  shown  by  cor- 
responding letters,  and  from  the  points  thus  obtained 
draw  lines  to  X',  cutting  the  line  A1  B'.  From  X'  as 
center  draw  arcs  indefinitely  from  each  of  the  points 
in  D1  C'.  From  any  convenient  point  upon  arc  a,  as 
D',  draw  a  line  to  X1,  which  will  form  one  side  of  the 
pattern  of  the  desired  envelope.  Take  between  the 
points  of  the  dividers  a  space  equal  to  that  used  in 
dividing  the  plan  G  F  E,  and,  placing  one  foot  of  the 


From  each  of  the  points  where  the  lines  b  b,  c  c, 
d  d,  etc.,  of  the  elevation  cross  the  line  of  the  opening 
project  lines  horizontally,  cutting  _hypothenuses  of 
corresponding  letter  in  the  diagram  of  triangles,  all  as 
shown  by  a\  b1  U,  c  c\  etc.  With  one  foot  of  the  com- 
passes at  X1,  bring  the  pencil  point  successively  to  the 
points  a1,  b',  b',  etc.,  and  draw  arcs  cutting  radial  lines 
in  the  pattern  of  corresponding  letter.  Then  a  line 
traced  through  the  points  thus  obtained  will  be  the  re- 
quired shape  of  the  opening  in  the  pattern. 


Pattern  Problems. 

PROBLEM    176. 

The  Pattern  for  a  Chimney  Top. 


327 


In  Fig.  566  are  shown  the  side  and  end  elevations 
and  the  plan  of  a  cliimncv  top.  A  B  C  D  of  the  plan 
represents  the  si/.o  of  the  article  at  the  bottom  to  fit 
the  chimney,  and  E  F  G  II  is  the  size  of  the  opening 


H' 

G2 

from  the  points  thus  obtained  draw  lines  to  C.  The 
next  step  is  to  construct  a  diagram  of  triangles  of 
which  the  lines  just  drawn  are  the  bases  and  of  which 
the  hight  of  the  article  is  the  altitude.  Assuming  0  C 
as  the  base  line  of  this  diagram,  place 
one  foot  of  the  compasses  at  C,  and, 
bringing  the  pencil  point  to  the  vari- 
ous points  in  F  G,  strike  arcs  cut- 
ting. 0  C,  as  shown.  At  right  angles 
to  0  C  erect  C  Q,  equal  in  hight  to 
J  H1  of  Fig.  566,  and  from  Q  draw 
lines  to  the  several  points  in  0  C. 
These  hypothenuses  will  then  repre- 
sent the  true  distances  from  C  to  the 
points  in  F  G.  From  Q  as  center, 
and  radii  equal  to  the  several  hy- 
potheuuses,  strike  arcs  indefinitely, 
as  shown  to  the  left.  From  any  con- 
venient point  on  arc  0,  as  G',  draw  a 
line  to  Q,  which  will  form  one  side  of 
the  pattern  of  the  rounded  corner.  Set  the  dividers  to 
the  space  used  in  stepping  off  the  arc  F  G,  and,  com- 


PLAN 
Fig.  566.— Flan  and  Elevations  of  Chimney  Top. 

in  the  top  to  fit  a  pipe  or  extension.  An  inspection 
of  the  drawings  will  show  that  the  article  consists  of 
two  flat  triangular  sides,  of  which  A  II  D  is  the  plan 
and  A1  H1  D1  the  elevation,  two  similar  triangles,  D  G 
C,  forming  the  ends,  and  four  corner  pieces,  of  which 
F  C  G  is  a  plan,  which  are  portions  of  an  oblique  cone. 
Further  inspection  of  the  plan  will  also  show  that  the 
entire  article  consists  of  four  symmetrical  quarters, 
therefore  in  Fig.  567  is  shown  a  quarter  plan  of  the 
same  in  which  corresponding  points  are  lettered  the 
same  as  in  Fig.  566,  from  which  the  pattern  for  one- 
quarter  of  the  article  is  obtained. 

The  quarter  circle  F  G  is  the  quarter  plan  of  an 
oblique  cone,  of  which  C  is  the  apex;  therefore  divide 
F  G  into  any  convenient  number  of  equal  parts,  and 


Fig.  567.— One-Quarter  Plan  and  Pattern  of  Chimney  Top.  . 

mencing  at  the  point  G1,  step  to  arc  1  and  from  that 
point  to  arc  2  and  so  on,  reaching  the  last  arc  in  the 


328 


Ttie  New  Metal  Worker  Pattern  Book. 


point  F'.  Trace  a  line  through  these  points,  as  shown 
from  F1  to  G',  and  draw  F'  Q,  which  will  complete  the 
pattern  of  the  corner  piece. 

From  C  set  off  on  0  C  the  distances  O  F  and  L  G,  as 
shown  by  M  and  N.  Draw  lines  from  these  points  to 
Q,  then  M  Q  and  N  Q  will  represent  the  true  distances 
shown  by  0  F  and  G  L  of  the  plan  or  J  II'  and  L  G"  of 
the  elevations  in  Fig.  5(56. 

With  Q  as  center,  and  C  L  as  radius,  describe  an 
arc,  L',  and  from  G1  as  center,  with  radius  equal  to  N 
Q,  intersect  the  arc,  as  shown,  thus  establishing  the 
point  L1.  Draw  G1  L1  and  L1  Q.  In  a  similar  manner, 
with  FX  as  center,  and  Q  M  as  radius,  describe  the  arc 
O',  and  from  Q  as  center,  with  a  radius  equal  to  C  0 
of  the  plan,  intersect  the  arc  at  the  point  O'.  Draw  Q 
O'  and  O1  F' ;  then  F1  O1  Q  L1  G'  will  form  the  pattern 
for  one  complete  quarter  of  the  chimney  top.  A  du- 
plicate of  this  pattern  may  be  added  to  it  if  desired, 


Fig.  568.— One-Half  Pattern  of  Chimney  Top. 

joining  the  two  upon   the   line  F'  O,  thus   forming  a 
pattern  for  one  ualf,  as  shown  in  Fig.  568. 


PROBLEM   177. 


The  Pattern  of  an  Article  with  Rectangular  Base  and  Round  Top. 


In  Fig.  569  are  shown  the  plan  and  elevations  of 
an  article  in  which  the  conditions  are  exactly  the  same 
as  in  the  preceding  problem.  The  article  here  shown 
differs  from  that  shown  in  Fig.  566  only  in  the  fact 
that  the  diameter  of  the  round  end  or  top  is  greater  than 
the  width  of  the  base,  while  in  Fig.  566  it  is  less,  but 
the  method  of  obtaining  the  pattern  is  exactly  the  same. 

In  this  case,  as  in  the  preceding  one,  the  article 
consists  of  four  flat  triangular  pieces  (two  ends  and  two 
sides)  and  four  equal  rounded  corners,  each  of  which  is 
a  quarter  of  an  oblique  cone.  As  the  entire  envelope 
consists  of  four  symmetrical  quarters,  one-quarter  of 
the  plan  0  P  N  J  has  been  reproduced  in  Fig.  570  from 
which  to  obtain  the  patterns  in  the  simplest  manner. 

Divide  J  I  of  plan  into  any  convenient  number  of 
equal  parts,  and  from  the  points  thus  obtained  draw 
lines  to  N,  which  represents  the  apex  of  an  inverted 
oblique  cone.  The  object  is  to  construct  triangles 
whose  altitudes  will  be  equal  to  the  straight  hight  of  the 
article,  and  whose  bases  will  be  equal  to  the  length  of 
lines  in  IJ  N  of  plan,  and  whose  hypothenuses  will  gi  ve 
the  distance  from  points  in  I  J  of  top  to  N  in  the  base. 


To  construct  this  diagram,  proceed  as  follows : 
From  N  of  Fig.  570  as  center,  and  radii  equal  to  the 
lengths  of  the  several  lines  drawn  to  N,  describe  arcs, 
cutting  any  straight  line,  as  N  W.  From  N  draw  N  n 
at  right  angles  to  N  W,  which  make  equal  to  llie  straight, 
hight  of  the  article,  and  from  the  points  in  N  W  draw 
lines  to  n.  With  n  as  center,  and  the  distances  from  N 
to  points  in  N  W  as  radii,  strike  arcs  as  shown.  From 
any  point,  as  i,  on  arc  1,  draw  a  line  to  n.  Set  the 
dividers  to  the  space  used  in  stepping  off  I  J  of  plan, 
and,  commencing  at  i,  step  from  arc  to  arc.  as  indicated 
by  the  small  figures,  reaching  the  last  in  the  point  7 
or/.  Draw  /  n,  thus  completing  the  pattern  for  part 
of  article  indicated  in  plan  by  I  N  J.  From  \  on  N 
W  set  off  the  distances  Q  J  and  I  P,  as  shown  by  the 
points  q'  and  t'.  Then  n  q'  and  n  t'  will  represent  re- 
spectively the  altitudes  of  the  ilat  triangular  pieces 
forming  the  sides  and  the  ends  of  the  article.  With 
N  Q  of  plan  as  radius,  and  n  of  pattern  as  center,  strike 
a  small  are  (</),  which  intersect  with  one  struck  from  j 
of  pattern  as  center,  and  n  q  of  diagram  as  radius,  thus 
establishing  the  point  q  of  pattern.  Draw  n  q  and  qj. 


Pattern  Problems, 


329 


With  P  N  of  plan  as  radius,  and  n  of  pattern  as  center, 


In  Fig.   571  ipqjis  a  duplicate  of  the  pattern 


strike  ;i  small  arc,  which  intersect  with  one  struck  from      indicated  by  the  same  letters  in  Fig.  570.      Below  this 
i  of  pattern  as  center,  and  n  t'  of  the  diagram  as  radius,      the  pattern  is  duplicated  once,  and  above  twice,  each 


N'  M' 

Fig.  569. — Plan  and  Elevations  of  an  Article  with  Rectangular  Base  and  Round  Top. 

1L 


fig.  570.— One-Quarter  Plan  and  Pattern  of  Article  Shoivn  in  Fig.  Sfiit. 


Fig.  571.— The  Entire  Pattern  of  Article  shown 
in  Fig.  569  in  One  Piece. 


thus  establishing  pointy  of  pattern.    Draw  ip&udpn, 
as  shown,  thus  completing  the  quarter  pattern. 


alternate  pattern  being  reversed,  thus  completing  the 
entire  pattern  in  one  piece. 


330 


Tl)e  New  Metal   Worker  Pattern  Book. 


PROBLEM  178. 

Pattern  for  an  Article  Forming  a  Transition  from  a  Rectangular  Base  to  an  Elliptical  Top. 


In  Fig.  572,  A  B  B'  A'  of  the  plan  shows  the 
rectangular  base  and  C  E  D  F  the  elliptical  top  of  an 
article,  the  sides  of  which  are  required  to  form  a  tran- 
sition between  the  two  outlines.  A"  C'  D'  B"  is  an 


obtaining  the  pattern  has,  however,  been  employed, 
not  because  it  is  better  but  for  the  sake  of  variety, 
leaving  the  reader  to  judge  which  method  is  the  more 
available  in  any  given  case.  Divide  E  D  into  any 

convenient  number  o  f 
equal  spaces,  as  shown  by 
the  small  figures,  and  from 
the  points  thus  obtained 
draw  lines  to  B.  These 
lines  will  form  the  bases 
of  a  series  of  triangles 
whose  common  altitude  is 
equal  to  the  hight  of  the 
article,  X  Y,  and  whose 
hypothenuses  when  ob- 
tained will  be  the  real  dis- 
tances from  B  in  the  base 


DIAGRAM    OF 
TRIANGLES 


A'  X  B' 

Fig.  572.— Plan  and  Elevation  of  Transition  Piece. 

end  elevation  of  the  same,  showing  its  vertical  hight  X 
Y.  An  inspection  of  the  plan  will  show  that  the  arti- 
cle consists  of  four  symmetrical  quarters,  and  that  that 
part  of  either  quarter  lying  between  the  curved  outline 
of  the  top  and  the  extreme  angle  of  the  bas"e,  as  the 
paft  E  D  B,  is  a  portion  of  the  envelope  of  an  oblique 
elliptical  cone,  of  which  E  D  is  the  base  and  B  the 
apex. 

The  conditions  here  given  are  exactly  the  same  as 
in  the  two  preceding  problems ;   a  different  method  of 


D  d 

Fifl.  57S. — Pattern  for  One-Quarter  of  Article  Shown  in  Fig.  572. 

to  the  points  assumed  in  the  curve  of  the  top.  To  con- 
struct such  a  diagram  of  triangles,  first  draw  any  line, 
as  L  M,  and  from  M  lay  off  the  distances  shown  by  solid 
lines  in  plan,  thus  making  M  1  equal  to  B  1,  M  2  equal 
to  B  2,  etc.  At  right  angles  to  L  M  draw  M  N,  in 
hight  equal  to  the  straight  hight  of  the  article,  as  shown 
by  X  Y  of  elevation,  and  connect  the  points  in  M  L  with 
N.  Also  set  off  the  distance  D  rffrom  M,  and  draw  N 
d.  If  E  d  was  different  in  length  from  D  d,  this  distance 
would  be  set  off  from  M  and  a  line  drawn  to  N. 


Pattern  Problems. 


331 


To  develop  the  pattern  first  draw  any  line,  as  E  B 
of  Fig.  573,  equal  in  length  to  N  1  of  the  diagram. 
With  B  of  pattern  as  center,  and  N  2  of  the  diagram 
of  triangles  as  radius,  describe  a  small  arc,  which  in- 
tersect with  another  arc  struck  from  E  of  pattern  as 
center  and  E  2  of  plan  as  radius,  thus  establishing 
point  ~2  of  pattern.  Proceed  in  this  manner,  using  the 
distance  between  points  in  plan  for  the  distance  be- 
tween similar  points  in  pattern,  and  the  hypothenuses 
of  the  triangles  in  the  diagram  in  Fig.  572  for  the  dis- 
tances to  be  set  off  from  B  of  pattern  on  lines  of  simi- 
lar number.  Through  the  points  thus  obtained  trace  a 
line,  as  shown  by  E  D.  With  B  of  pattern  as  center, 
and  B  d  of  plan  as  radius,  strike  a  small  arc,  which  in- 


tersect with  another  struck  from  D  of  patterns  as  cen- 
ter, and  N  d  of  diagram  as  radius,  thus  establishing  the 
point  <l  of  the  pattern.  Draw  D  d  and  d  B.  With  B 
of  the  pattern  as  center,  and  B  e  of  the  plan  as  radius, 
strike  a  small  arc,  which  intersect  with  another  struck 
from  E  as  center,  with  a  radius  equal  to  N  d  of  the 
diagram  of  triangles.  Draw  E  e  and  e-B ;  then  E  e  B 
d  D  will  be  the  pattern  for  one-quarter  of  the  article. 

In  performing  the  work  of  development  of  the  pat- 
tern it  will  be  found  convenient  as  well  as  more  accu- 
rate to  use  two  pairs  of  compasses,  one  of  which  should 
remain  set  to  the  space  used  in  dividing  the  curve  E  D  of 
the  plan,  while  the  other  may  be  changed  to  the  varying 
lengths  of  the  hypothenuses  in  the  diagram  of  triangles. 


PROBLEM  179. 

Pattern  for  an  Article  Forming  a  Transition  from  a  Rectangular  Base  to  a  Round  Top,  the  Top  Not 

Being  Centrally  Placed  Over  the  Base. 


In  Fig.  574,  F  G  II  J  of  the  plan  represents  the 
bottom  of  the  article  and  A  B  D  E  the  top.     Below 


J' 


FRONT  ELEVATION 
Fig.  574.— Elevations  and  Plan  of  an  Irregular  Transition  Piece. 


the  plan  is  projected  a  front  elevation  and  at  the  right 
a  side  elevation,  like  points  in  all  the  views  being 
lettered  the  same.  An  inspection  of 
the  drawing  will  show  that  each  side 
of  the  article  consists  of  a  triangular 
piece  whose  base  is  a  side  of  the  rect- 
angle and  whose  vertex  lies  at  a  point 
in  the  circle  of  the  top,  the  four 
vertices  marking  the  division  of  the 
circle  into  quarters,  and  four  quarters 
of  inverted  oblique  cones  whose  bases 
are  the  quarter  circles  of  the  top  and 
whose  apices  lie  at  the  corners  of  the 
rectangle.  A  comparison  between 
this  figure  and  the  one  shown  in 
Problem  177  will  show  that  the  con- 
ditions existing  in  either  one  of  the 
corner  pieces  in  this  case  are  exactly 
the  same  as  in  the  former  problem, 
but  that  while  in  Problem  177  the 
four  corners  are  alike,  in  the  present 
instance  the  four  corners  are  all  differ- 
ent, and  that  therefore  the  pattern  for 
each  corner  piece,  as  well  as  that  for 
each  of  the  flat  sides,  must  be  ob- 
tained at  a  separate  operation,  all  being 
finally  united  into  one  pattern. 

Divide  the  plan  of  the  top  A  B 
D  E  into  any  convenient  number  of 
equal  spaces  in  such  a  manner  that 


332 


New  Metal   Worker  Pattern  Book. 


each  quarter  of  the  circle  shall  contain  the  same 
number  of  spaces  and  from  the  points  of  division  in 
each  quarter  draw  lines  to  the  adjacent  corner  of  the 
rectangle  of  the  base,  all  as  shown  in  Fig.  575.  Thus 
lines  from  the  points  in  P]  D  are  drawn  to  H  and 
lines  from  points  in  D  B  are  drawn  to  G,  etc. 

The  next  operation  will  be  to  construct  the  four 
diagrams  of  triangles  (one  for  each  corner  piece) 
shown  in  Fig.  576,  of  which  these  lines  are  the  bases. 
Accordingly  lay  off  at  any  convenient  place  the  line 
H  L,  Fig.  576,  equal  to  the  straight  hight  of  article, 
as  shown  by  J'  X  in  Fig.  574.  From  the  point  H, 
and  at  right  angles  to  L  H,  draw  the  line  H  M,  and. 
measuring  from  H,  set  off  the  length  of  lines  in  E  D 


Fig.  575.— Plan  of  Irregular  Transition  Piece  with  Surface 
Divided  into  Triangles. 


H  of  the  plan,  Fig.  575.  Thus  H  1  is  made  equal  to 
H  E  of  the  plan,  H  2  is  made  equal  to  the  distance 
H  2  of  the  plan,  and  H  3  is  equal  to  distance  H  3  of 
the  plan,  etc.  From  the  points  thus  established  in 
M  H  draw  lines  to  L,  as  shown.  Then  the  hypoth- 
enuses  L  1,  L  2,  L  3,  etc.,  will  correspond  to  the  width 
of  the  pattern,  measured  between  points  in  E  D  of  top 
and  H  in  the  base. 

The  triangles  for  the  corner  piece  D  G  B  are  con- 
structed in  the  same  general  manner.  N  G  corre- 
sponds to  the  hight  J'  X  of  the  elevation.  0  G  is 
drawn  at  right  angles  to  N  G,  on  0  G  are  set  off  the 
lengths  of  lines  in  D  G  B  of  the  plan,  and  from  the 
points  thus  obtained  lines  are  drawn  to  N.  Thus  G  5 


of  the  diagram  is  equal  to  G  5  of  the  plan,  G  6  of  the 
diagram  is  equal  to  G  6  of  the  plan,  etc.  The  tri- 
angles in  P  Q  F  correspond  with  the  lines  in  A  F  B  of 
the  plan,  as  do  those  in  S  R  J  with  the  lines  A 
J  E  in  the  plan.  Before  commencing-  to  describe  the 
pattern  the  seam  or  joint  may,  for  convenience,  be 
located  at  K  E  of  the  plan.  The  real  length  of  the  line 
Iv  K  of  the  plan  is  given  by  II"  E"  in  the  side  eleva- 
tion. Fig.  574,  or  the  distance  E  K  can  be  set  olT.  as 
shown  by  IT  K  i:i  Fig.  .">7»'>.  The  dotted  lino  K  L 
will  then  be  the  distance  from  K  in  the  base  to  E  in 
the  top. 

As  it  is  necessary  in  obtaining  the  pattern  for  the 
entire  envelope  that  the  patterns  of  the  parts  shall  suc- 


9  1O  11  13  12        F 

Fig.  576. — Diagrams  of  Triangles  Obtained  from  fig.  575. 

ceed  one  another  in  the  order  in  which  they  occur  in 
the  plan,  the  method  of  development  here  adopted  is 
that  of  constructing  separately  each  small  triangle,  as 
in  the  preceding  problem,  instead  of  by  means  of  a 
number  of  arcs,  as  in  Problem  177  and  others  preced- 
ing it.  To  begin,  then,  with  the  pattern  of  the  part 
corresponding  to  F  B  G  of  the  plan.  The  length  F  G 
of  the  pattern,  in  Fig.  577,  is  established  by  the  length 
F  G  of  the  plan  in  Fig.  575.  With  F  of  pattern  as 
center,  and  P  9  of  Fig.  576  as  radius,  describe  an  arc, 
B,  which  intersect  with  one  struck  from  G  of  the  pat- 
tern as  center,  and  N  9  of  Fig.  576  as  radius,  thus  es- 
tablishing the  point  B  of  the  pattern.  Then  F  B  G  is 


Pattern  Problems. 


333 


tlic  pattern  for  that  part  of  the  .article  shown  by  F  B 
G  of  tlic  plan.  From  G  of  pattern  as  center,  with 
radii  corresponding  to  the  hypothenuses  of  the  tri- 
angles shown  in  0  N  G  of  Fig.  576.  strike  the  arcs 
shown.  Thus  G  S  of  the  pattern  is  equal  to.N  8,  G  7 
of  the  pattern  is  equal  to  N  7,  etc.  With  the  dividers 
set  to  the  same  space  used  in  stepping  off  the  plan, 
with  B  or  9  of  the  pattern  as  center,  strike  a  small  arc 
intersecting  arc  S  previously  drawn,  thus  locating  the 
point  S.  From  S  as  center  intersect  arc  7,  and  so 
continue,  locating  the  points  <>  and  ">.  Through  the 
points  thus  obtained  can  be  traced  the  line  B  D.  Then 


With  the  dividers  set  to  the  same  space  as  was  used  in 
stepping  off  the  plan,  and  commencing  at  5,  intersect 
each  succeeding  arc  from  the  point  obtained  in  the  one 
before  it,  as  shown  by  the  figures  4,  3,  2,  1.  Trace  a 
line  through  the  points  thus  obtained,  and  connect  E' 
H,  as  shown.  Then  E'D  II  is  the  pattern  for  that  part 
of  the  article  shown  on  plan  by  E  D  H.  With  II  of 
pattern  as  center,  and  II  K  of  plan  as  radius,  describe 
a  small  arc,  which  intersect  with  one  struck  from  E' 
of  pattern  as  center,  and  L  K  of  Fig.  576,  or  what  is 
the  same,  E"  II"  of  Fig.  574,  as  radius,  thus  establish- 
ing the  point  K'  of  the  pattern.  Connect  H  K'  and 


/ 

\                   y 

/ 

\/ 

./ 

\/ 

F 

C 

Fig.  577. — Pattern  uf  Transition  Piece,  Shown  in  Fig.  574. 


(i  B  1)  is  the  pattern  for  that  part  of  the  article  shown 
on  the  plan  by  G  B  D. 

With  G  of  pattern  as  center,  and  G  H  of  plan  as 
radius,  strike  a  small  arc,  H,  which  intersect  with  one 
struck  from  D  of  pattern  as  center  and  L  M  of  Fig. 
576  as  radius,  thus  establishing  the  point  H  of  pattern. 
Connect  G  H  and  H  D,  as  shown.  Then  G  H  D  is  the 
pattern  for  that  part  of  the  article  shown  in  plan  by  G 
II  D.  With  H  of  pattern  as  center,  and  the  hypoth- 
enuses of  triangles  in  M  L  H  of  Fig.  576  as  radii,  strike 
arcs,  as  shown,  making  H  4,  II  3,  H  2,  H  1  of  pattern 
equal  to  L  4,  L  3.  L  2,  L  1  of  the  diagram  of  triangles. 


K'  E',  as  shown,  which  gives  the  pattern  for  that  part 
of  the, article  shown  on  plan  by  H  K  E. 

The  radii  for  striking  the  arcs  in  A  F  B  of  the 
pattern  are  found  in  O  P  F  of  Fig.  576.  The  length 
F  J  of  pattern  is  established  by  the  length  F  J  of  the 
plan.  The  radii  for  striking  the  arcs  in  A  J  E  of  pat- 
tern are  found  in  S  li  J  of  Fig.  576.  J  K  of  the  pat- 
tern corresponds  with  J  K  of  the  plan,  and  E  K  of  the 
pattern  corresponds  with  L  K  of  Fig.  576.  Thus  E  A 
B  D  E'  of  the  pattern  is  the  stretchout  of  E  A  B  D  of 
the  plan  of  the  top,  as  K  J  F  G  H  K'  of  the  pattern  is 
the  stretchout  of  K  J  F  G  H  of  the  plan  of  the  base. 


PROBLEM    180. 

The  Pattern  for  a  Collar  Round  at  the  Top  and  Square  at  the  Bottom,  to  Fit  Around  a  Pipe  Passing 

through  an  Inclined  Roof. 


Let  A  B  D  C  of  Fig.  578  represent  the  side  eleva- 
tion of  the  pipe  and  C  D  E  F  the  side  view  of  the  col- 
lar, titting  against  the  pitch  of  the  roof  shown  by  G  H. 


Construct  a  plan  below  the  elevation,  as  shown,  making 
J  K  M  L  the  plan  view  of  tire  pipe  and  N  0  P  R  the 
plan  view  of  the  collar  on  a  horizontal  line,  giving  the 


334 


Tlie  New  Metal    Worker  Pattern  Book. 


collar  an  equal  projection  at  the  bottom  on  the  four 
sides,  as  shown.  Through  the  center  point  X  in  plan 
draw  a  line  parallel  to  N  R,  intersecting  the  circle  at 
K  and  L  ;  likewise  through  the  center  X,  and 
parallel  to  0  N,  intersect  the  circle  at  J  and  M.  From 
J  and  K  draw  lines  to  the  corner  N ;  likewise  from  .1 
and  L,  L  and  M,  and  M  and  K,  draw  lines  to  the  cor- 
ners R,  P  and  0.  It  will  be  seen  that  by  this  opera- 


the  bases  of  a  series  of  right  angled  triangles,  whose 
hypothenuses  will  give  the  correct  distances  across  the 
pattern  of  the  collar.  To  construct  these  triangles 
proceed  as  follows :  Upon  C  Y  extended  assume  any 
point,  as  S,  at  which  erect  the  perpendicular  ST,  equal 
in  hight  to  the  cone  C  Y  F,  as  shown  bv  the  dotted 
line  from  F.  From  S  on  S  C  set  off  the  lengths  of  the 
several  lines  in  K  N  J  of  the  plan,  as  shown  by  1',  2', 


,  2'  X 

S                                3'    4'     5'  ^G 



SIDE  ELEVATION 
C                      Y' 

D 

5'      6' 
9'        8'  7' 


DIAGRAM  OF 
TRIANGLES  IN  J  K  N 


Fig.  578.— Plan  and  Side  Elevation  of  a  Collar  to  Fit  Around  a  Pipe  Pa  siny  throu  h  an  Inclined  Roof. 


tion  the  collar  has  been  divided  in  such  a  manner  that 
the  four  corner  pieces  are  portions  of  oblique  cones 
whose  apices  lay  at  the  corners  of  the  collar,  while  the 
side  pieces  between  are  simply  flat  triangular  pieces  of 
metal.  The  dotted  lines  connecting  the  plan  with  the 
elevation  show  corresponding  points  in  the  two  views. 
Divide  the  quarter  circles  K  J  and  J  L  into  any 
convenient  number  of  equal  spaces,  as  shown  by  the 
small  figures,  and  from  points  on  each  draw  lines  to 
the  corners  N  and  R.  Then  will  these  lines  represent 


3',  etc.,  and  from  these  points  draw  lines  to  T.  In 
the  same  manner  construct  the  diagram  of  triangles 
shown  at  the  right.  At  U  upon  the  line  Y  D  extended 
erect  the  perpendicular  U  V,  equal  in  hight.to  the  conn 
Y  D  E,  as  shown  by  the  dotted  line  E  V.  From  I' 
on  U  D  set  off  the  lengths  of  the  lines  in  J  R  L  and 
draw  the  hypothenuses,  as  shown. 

To  develop  the  pattern,  first  draw  any  horizontal 
line,  as  A  A1  of  Fig.  579,  equal  in  length  to  P  R  in 
plan  of  Fig.  578.  With  A  and  A'  as  centers,  and  the 


Pattern  Problems. 


335 


hypothenuse  V  9'  of  Fig.  578  as  radius,  describe  arcs 
intersecting  each  other  at  9.  Now,  with  9  of  the  pat- 
tern us  center,  and  9  8  of  the  plan  as  radius,  describe 


A  A 

Fig.  579.— Pattern  of  Collar  Shown  in  Fig.  578. 

the  arc  S ;  then  with  V  8'  of  Fig.  578  as  radius,  and  A 
of  the  pattern  as  center,  describe  an  arc,  intersecting 
the  arc  previously  drawn,  thus  establishing  the  point  8. 
Proceed  in  this  manner,  using  alternately  first  the 
divisions  on  the  quarter  circle  L  J  in  plan,  then  the 
livpothenuses  of  the  triangles  whose  bases  are  shown  by 
the  lines  in  J  L  K,  until  the  point  5  in  pattern  has 


been  obtained.     Draw  a  line  from  5  to  A  in  Fig.  579. 
Then   with   A  as   center,  and  E  F  in  side  elevation, 
Fig.  578,   as  radius,   describe  an  arc,  shown  at  C  of 
Fig.  579,  and  with  5  of  the  pat- 
tern as  center,  and  the  hypothe- 
nuse T  5'  of  Fig.  578  as  radius, 
describe  an  arc    intersecting    the 
previous  arc  at  C.     Draw  a  line 
from  5  to  C.     Now    proceed   as 
above  described,  using  alternately 
first   the   spaces   on    the    quarter 
circle  in   J  K  in  plan,  then  the 
hypothenuses    of     the     triangles 
whose  bases  are  shown  in  J  K  N 
in  plan,  until  the  point  1  in  pat- 
tern  has    been    obtained.     Then 
with  C  of  the  pattern  as  center, 
and  W  N  of  the  plan  as  radius, 
describe  an  arc,  as  shown  at  D, 
and  with  F  C  in  side  elevation  as 
radius,  and  1    of   the   pattern  as 
center,     describe    an   arc,     inter- 
secting  the    arc    previously    de- 
scribed at  D.     Draw  the  lines  1 
D,  D  C,   C  A,  and  through    the 
intersections    of   the   arcs    trace   a  line,   shown    from 
1  to  9  on  pattern.     This  will  complete  one-half  the 
pattern.       The    entire  pattern  may  be  completed   by 
duplicating  the  part  1  5  9  A  C  D  and  adding  the  same 
to  that   already  obtained  in  such  a  manner  that  the 
side  9  A  will  coincide  with  9  A1,  as  shown  by  9  5'  1' 
D1  C'  A1. 


PROBLEM    181. 


The  Pattern  for  a  Flaring  Article  Round  at  the  Base  and  Square  at  the  Top. 


The  shape  shown  in  Fig.  580  differs  from  that 
treated  in  Problem  176  principally  in  the  fact  that  the 
round  end  is  larger  than  the  rectangular  end  instead  of 
smaller  as  in  Fig.  566 ;  the  conditions  involved  are, 
however,  exactly  the  same  as  in  the  other  problem 
Mini  consequently  the  method  of  obtaining  t\\e  pattern 
must  be  similar.  F  G  II  J,  in  Fig.  580,  represents 
the  plan  of  the  base,  K  L  M  N  that  of  the  top,  and 
A  B  C  E  the  elevation  of  a  side  of  the  article. 


Through  0,  the  center  of  the  circle  of  the  base,  draw 
the  diameters  G  J  and  F  H  parallel  to  the  sides  of  the 
top.  From  the  four  points  thus  obtained  in  the  cir- 
cumference of  the  base  draw  lines  to  the  angles  of  the 
top,  as  shown  by  G  M  and  H  M,  H  N  and  J  N,  etc. 
It  will  be  seen  from  this  that  the  envelope  of  the  ar- 
ticle consists  of  four  flat  triangles,  of  which  L  G  M  is 
a  plan  and  B  D  C  the  elevation,  and  four  rounded 
corners,  either  one  of  which,  as  J  N  H,  is  a  portion  of 


336 


The  Xcw  Metal 


/'<///,,•,* 


an  obli«[uc  cone  of  which  .J  II  is  the  base  and  N 

apc\. 


the 


Fig,  580.— Plan  and  Elevation  of  a  Flaring  Article,  Round  at 
the  Base  and  Square  at  the  Top. 

To  obtain  the  pattern  first  divide  the  quarter  plan 
of  base  J  H  into  any  convenient  number  of  parts,  as 
indicated  by  the  small  figures,  and  con- 
nect these  points  with  N,  as  shown. 
To  obtain  the  distance  from  points  in 
J  H  of  base  to  N  of  top  it  will  be 
necessary  to  construct  the  diagram  of 
triangles  shown  in  Fig.  581.  Draw  any 
line,  as  R  P,  in  length  equal  to  the 
hight  of  the  article,  as  shown  by  S  C  in 
Fig.  580.  At  right  angles  to  R  P  draw 
P  Q,  and  on  P  Q  lay  off  the  lengths  of 
lines  in  J  H  N.  Thus  make  P  1  equal 
to  N  1  of  the  plan,  P  2  equal  to  N  2, 
etc.  Connect  the  points  in  P  Q  with 
R.  The  hypothenuses  thus  obtained 


give  the  true  distances  from  the  points   in   the  base  to 
N  in  the  top. 

.  From  any  convenient  point,  as  N  in  Fig.  5S2,  as 
center,  with  radius  R  1  of  Fig.  581,  describe  an  arc, 
as  shown  by  1  7.  In  like  manner,  with  radii  R  2, 
R  3  and  R  -4  of  Fig.  581,  describe  arcs,  as  shown. 
Draw  a  straight  line  from  N  to  anv  convenient  point 
upon  the  arc  1  7,  as  shown  by  X  11.  Set  the  dividers 
to  the  space  used  in  stepping  oil'  the  plan  of  the 
base  and,  starting  with  H,  lay  off  the  stretchout, 


43     2 
5    6 


Fig.  S81. — Jjiayram  of  THanyles  Used  in  Obtaining  Pattern  of 
Article  Shown  in  Fig.  aSO. 

stepping  from  arc  to  arc,  as  shown.  A  line  traced 
through  these  points  will  form  the  pattern  for  as  much 
of  the  article  as  shown  bv  J  X  II  of  the  plan.  With 
X  of  pattern  as  center,  and  X  K  of  plan  or  B  C  of  ele- 
vation as  radius,  describe  a  small  arc,  K,  which  inter- 
sect with  an  arc  struck  from  J  of  pattern  as  center  and 
J  X  as  radius.  Connect  J  K  and  K  X,  which  com- 
pletes the  pattern  for  J  K  X  of  the  plan.  .1  K  !•'  of 
pattern  is  the  same  as  J  X  H,  and  can  be  obtained  in 
the  same  manner,  or  by  any  convenient  means  of  du- 
plication. With  X  as  center,  and  X  W  of  plan  as 
radius,  describe  a  small  arc,  which  intersect  with 

W 


Fig.  5H2.— One-Half  of  Pattern  of  Article  Shown  in  Fig.  580. 


Pattern  Problems. 


337 


struck  from  II  as  center,  and  E  G  of  elevation  as 
radius.  Connect  II  W  and  W  N,  thus  producing  the 
part  of  pattern  corresponding  to  N  W  II  of  the  plan. 


F  K  V  of  pattern  is  obtained  in  a  similar  manner. 
Then  V  K  N  W  H  J  F  is  the  pattern  for  one-half  of 
article. 


PROBLEM  182. 

Pattern  for  an  Article  Rectangular  at  One  End  and  Round  at  the  Other,  the  Plane  of  the  Round  End 

Not  Being;  Parallel  to  that  of  the  Rectangular  End. 


In  Fig.  5S:-!  arc  shown  front  and  side  views  and 
I  Ian  of  an  article  forming  a  transition  between  a  rect- 
angular pipe  at  one  end  and  a  round  pipe  at  the  other, 
and  forming  at  the  same  time  an  angle  between  the 


Fig.  58S.— Front  and  Side  Views  and  Plan  of  an  Article  Forming  a  Transition  Between 
a  Rectangular  Pipe  and  a  Round  Pipe,  at  an  Angle. 


two  pipes.  A  B  F  C  of  the  front  view  shows  the  size 
of  the  rectangular  pipe,  while  G  B  II  D  shows  the 
opening  to  receive  the  round  pipe.  In  the  side  view 
a  >•  shows  the  vertical  rectangular  end,  and  I  d  shows 
the  angle  at  which  the  round  end  is  placed,  b  I  d  being 
a  half  profile  of  the  round  end.  As  will  be  seen  by  an 
inspection  of  the  front  view,  each  quarter  of  the  circu- 
lar opening  is  treated  as  the  base  of  a  portion  of  a 
scalene  cone  whose  apex  is  in  the  adjacent  angle  of  the 
rectangle,  the  intermediate  surfaces  being  flat  triangu- 


lar pieces.  Thus  B  G  and  G  D  are  the  quarter  bases 
of  scalene  cones  whose  apices  are  respectively  at  A 
and  C ;  A  B  E  and  CDF  are  triangles  whose  altitudes 
or  profiles  are  shown  respectively  by  a  b  and  e  d  of  the 
side  view ;  and  A  G  C  is  a  triangle 
whose  profile  appears  at  o  u  in  the 
plan. 

Divide  each  quarter  of  the  pro- 
file b  I  and  I  d  of  Fig.  583  into  any 
number  of  equal  spaces,  as  shown 
by  the  small  figures ;  also  draw  a 
duplicate  of  this  half  profile  in 
proper  relation  to  the  plan,  as  shown 
by  m  g  x,  which  divide  as  before, 
numbering  the  points  in  each  to 
correspond,  as  shown.  From  the 
points  in  b  I  p  drop  lines  at  right 
angles  to  b  d,  cutting  the  same. 
From  the  points  in  m  g  x  carry 
lines  indefinitely  to  the  left  parallel 
to  the  center  line  gf,  and  intersect 
them  by  lines  of  corresponding 
number  erected  vertically  from  the 
points  in  I  d.  A  line  traced  through 
the  points  of  intersection  will  give 
a  correct  plan  view  of  the  opening 
in  the  round  end.  To  avoid  con- 
fusion of  lines  the  intersections 
from  the  points  between  b  and  h  or 
the  upper  half  of  the  opening  are 
shown  only  in  the  near  or  lower 
half  of  the  plan  from  t  to  w,  while  the  points  belong- 
ing to  the  lower  half  (h  to  d)  are  shown  only  in  the 
further  half  of  the  plan  from  p  to  q. 

From  each  of  the  points  in  p  q  of  the  plan  draw 
lines  to  s,  which  is  the  projection  of  e  of  the  side  view 
or  apex  of  the  cone  in  the  lower  half,  and  from  the 
points  in  t  u  draw  lines  to  o,  the  apex  of  the  cone  of 
the  upper  half  of  the  article.  These  lines  represent 
only  the  horizontal  distances  from  s  and  o  to  the  points 
in  the  opening  t  u  q  p  of  the  plan  or  B  G  D  of  the  front 


338 


The  X'-tr  Metal    Worker  J\tttent  Bouk. 


view.  To  ascertain  the  real  distances  between  these 
points  it  will  be  necessary  to  first  ascertain  their  ver- 
tical hights  from  an  assumed  horizontal  plane  and  then 
to  construct  from  these  measurements  a  series  of  right 
angled  triangles  whose  hypothenuses  will  give  the  de- 
sired distances. 

From  the  points  in  I  h  of  the  side  view  drop  lines 
vertically,  cutting  a  horizontal  line  drawn  from  a,  as 
shown  between  v  andj;  and  from  the  points  in  h  d  drop 
lines  to  w  z  drawn  horizontally  from  e.  To  construct 
the  triangles  required  in  the  top  part,  first  draw  the 
right  angle  R  0  K,  as  shown  in  Fig.  584,  and  from  0 
on  0  R  set  off  the  length  of  lines  in  b  hj  v  of  side  view, 
as  indicated  by  the  small  figures.  From  0  on  O  K 
set  off  the  length  of  lines  in  o  t  u  of  plan  of  top,  also 
as  indicated  by  the  small  figures.  Connect  the  points 
in  0  R  with  those  of  similar  number  in  0  K,  as  shown. 
To  obtain  the  triangles  required  for  the  bottom  part, 
proceed  in  a  similar  manner.  Draw  the  right  angle 
W  S  L,  as  shown  in  Fig.  585.  From  S  on  S  W  set  off 
the  length  of  lines  in  h  d  z  w  of  side  view,  as  indicated 
by  the  small  figures.  From  S  on  S  L  set  off  the  length 
of  lines  in  s  p  q  of  plan  of  top,  also  as  indicated  by  the 
small  figures.  Connect  the  points  in  S  W  with  those 
of  similar  number  in  S  L,  as  shown. 

For  the  pattern  proceed  as  shown  in  Fig.  586. 
Draw  the  line  0  0',  in  length  equal  to  A  B  of  front 
view  or  o  k  of  plan.  Bisect  0  0'  in  C,  and  erect  the 


D  of  pattern  as  center,  and  b  2  in  b  I  of  profile  as  radius, 
thus  establishing  point  2  of  pattern.      Proceed  in  this 


Fig.  584.— Diagram  of  Triangles  in  Top  Half. 

manner,  using  the  length  of  lines  in  R  0  K  for  dis- 
tances from  0  of  pattern,  and  the  stretchout  between 
points  in  b  I  of  profile  of  side  view  for  *he  distance  be- 


Fiij.  585. — Diagram  of  Triangles  in  Bottom  Half. 

tween  points  in  D  G  of  pattern;    then  draw  D  G  ami 
G  O.      With  point  G  of  pattern  as  center,  and   5   5'  in 


Fig.  586.— Pattern  for  Transition  Piece  Shovm  in  Fig.  583. 


perpendicular  C  D,  in  length  equal  to  a  b  of  side  view, 
and  draw  O  D,  D  O'.  These  lines  are  equal  in  length 
to  R  K  of  first  diagram  of  triangles.  With  0  of  pat- 
tern as  center,  and  2  2'  in  R  0  K  as  radius,  describe  a 
small  are,  2,  which  intersect  with  one  struck  from  point 


W  S  L  of  triangles  as  radius,  strike  a  small  arc,  E, 
which  intersect  with  one  struck  from  point  0  of  pat- 
tern as  center  and  a  e  of  side  view  as  radius,  thus  estab- 
lishing point  E  of  pattern.  Draw  G  E  and  E  O.  With 
point  E  of  pattern  as  center,  and  6  6'  of  triangle  in 


Pattern  Problems. 


339 


W  S  L  as  radius,  strike  a  small  arc,  6,  which  intersect 
with  one  struck  from  point  G  of  pattern  as  center,  and 
I  6  of  profile  as  radius,  thus  establishing  point  6  of  pat- 
tern. Proceed  in  this  manner,  using  the  length  of 
lines  in  W  S  L  as  distance  from  E  of  pattern,  and  the 
stretchout  between  points  in  I  d  of  profile  of  side  view 
for  the  distance  between  points  in  G  H  of  pattern,  and 
draw  G  II  and  II  E.  With  point  H  of  pattern  as  cen- 
ter, and  e  d  of  side  view  as  radius,  strike  a  small  arc, 
C,  which  intersect  with  one  struck  from  point  E  of 
pattern  as  center,  and  o/of  plan,  or  C  0  of  pattern,  as 


radius,  thus  establishing  point  C  of  pattern ;  then  draw 
H  C  and  C  E.  From  E  and  C  erect  the  perpendicu- 
lars E  R  and  C  F,  in  length  equal  to  c  e  of  side  view, 
and  draw  F  R.  With  0  of  pattern  as  center,  and  a  c 
of  the  side  view  as  radius,  strike  a  small  arc,  which 
intersect  with  one  struck  from  E  of  pattern  as  center 
and  e  c  of  the  side  view  as  radius,  thus  establishing 
point  K  of  pattern,  and  draw  O  K  and  K  E.  Then 
DGHFREKC  represents  the  half  pattern  of  article. 
The  other  half  can  be  obtained  in  the  same  manner  or 
by  duplication,  as  may  be  found  convenient. 


PROBLEM   183. 

The  Pattern  for  a  Flaring  Article,  Round  at  Top  and  Bottom,  the  Top  Being:  Placed  to  One  Side  of 

the  Center,  as  Seen  in  Plan. 


In  Fig.   587,  the  elevation  of  the  article  is  shown 
liv  A  B  D  C,   below  which  is  drawn  the  plan  of  the 


Fiy.  687.— Plan  and  Elevation  of  an  Irregular  Flaring  Article,  with 
Lines  of  Triangulation  Shown  in  the  Plan. 

same,  corresponding  parts  in  each  being  connected  by 
the  vertical  dotted  lines.     There  are  two  methods  of 


triangulation  available  in  the  solution  of  this  problem 
only  one  of  which  is  given  in  this  connection. 

Divide  one-half  of  the  circle  representing  the  base 
of  the  article  into  any  convenient  number  of  spaces,  as 
indicated  by  the  small  figures,  1,  2,  3,  etc.  In  like 
manner  divide  the  inner  circle,  which  represents  the 
top,  into  the  same  number  of  spaces,  as  indicated  by 
1',  2',  3',  etc.  Between  the  points  of  like  numbers  in 
these  two  circles,  as  for  example  between  2  and  2',  3  and 
3',  etc.,  draw  lines,  as  shown;  also  connect  the  points  in 
the  inner  circle  with  points  in  the  outer  circle  of  the 
next  higher  number,  as  indicated  by  the  dotted  lines. 
Thus,  connect  1'  with  2,  and  2'  with  3,  and  so  on,  as 
shown.  These  lines  just  drawn  across  the  plan  are 
the  bases  of  a  number  of  right  angled  triangles  whose 
altitudes  are  equal  to  the  hight  of  the  article,  and 
whose  hypothenuses,  when  drawn,  will  give  the  cor- 
rect distances  across  the  pattern,  or  envelope  of  the 
article,  between  the  points  in  the  top  and  those  in  the 
bottom  in  the  direction  indicated  by  the  lines  of  the 
plan.  The  triangles  having  the  solid  lines  of  the  plan 
as  their  bases  are  shown  in  Fig.  588,  while  those  con- 
structed upon  the  dotted  lines  are  shown  in  Fig.  589, 
and  are  obtained  in  the  following  manner : 

At  any  convenient  point  erect  a  perpendicular, 
E  F,  Fig.  588,  which  in  length  make  equal  to  the 
straight  hight  of  the  article,  as  shown  in  the  elevation. 
From  F  at  right  angles  set  off  a  base  line  of  indefinite 
length.  On  this  line,  measuring  from  F,  set  off  lengths 
equal  to  the  several  solid  lines  in  the  plan.  For 
example,  make  the  space  F  10  equal  to  the  length  10 


3-iO 


Tlie  New  Metal    Worker  Pattern  Book. 


10'  in  the  plan,  and  the  space  F  9  equal  9  9'  in  the 
plan,  and  so  on,  until  F  1  is  sot  off  equal  to  1  1'  in 
the  plan.  From  the  points  thus  established  in  the  bast- 
line  draw  lines  to  the  point  E.  The  triangles  thus 
constructed  will  represent,  sections  through  the  arti- 
cle on  the  solid  lines  in  the  plan.  In  other  words, 
the  several  hypothenuses  of  the  triangles  shown  in 
Fig.  588  are  equal  in  length  to  lines  measured  at  cor- 
responding points  on  the  surface  of  the  completed  ar- 
ticle. 

In  like  manner  construct  the  triangles  shown  in 
Fig.  589,  representing  measurements  taken  on  the 
dotted  lines  shown  in  the  plan.  Draw  the  perpen- 
dicular K  G,  equal  in  length  to  the  straight  hight  of 
the  article.  From  K  lay  off  a  horizontal  base  line  in- 
definite in  length,  drawing  it  at  right  angles  to  K  G. 
From  K  set-off  lengths  equal  to  the  dotted  lines  in  the 
plan — that  is,  making  the  distance  K  10  equal  to  the 


Fig.  588. — Diagram  of  Triangles 
Based  upon  the  Solid  Lines  of 
the  Plan  in  Fig.  587. 


Fig.  589. — D'agram  of  Triangles 
Based  upon  the  Dotted  Lines  of 
the  Plan  in  Fig.  587. 


distance  9'  10  in  the  plan,  and  K  9  equal  to  the  dis- 
tance 8'  9  in  the  plan,  and  so  on  until  K  2  is 
made  equal  to  the  distance  1'  2  in  the  plan.  From 
the  points  thus  established  in  the  base  line  draw 
lines  to  the  point  G.  Then  the  hypothenuses  of  the 
triangles  thus  constructed  will  equal  measurements 
along  the  surface  of  the  completed  article  at  points 
corresponding  to  the  dotted  lines  in  the  plan.  With 
distances  thus  established  upon  the  surface  of  the  ar- 
ticle, and  with  the  stretchout  of  the  required  pattern 
determined  at  both  top  and  bottom,  it  is  easy  to  lay 
out  the  pattern  upon  the  general  plan  of  constructing 
a  triangle  when  the  three  sides  are  given. 

The  development  of  the  pattern  can  be  begun  at 
either  end  according  to  convenience,  and  the  operation 
is  conducted  as  follows:  Assume  any  line,  as  1  1'  of 
Fig.  590,  which  make  equal  in  length  to  A  B  of  the 
elevation,  or,  what  is  the  same  thing,  equal  to  E  1  of 


Fig.  5S8,  which  is  one  side  of  the  first  triangle.  The 
other  two  sides  are  respectively  the  distance  1  2  of 
the  plan  and  the  liypothenuse  of  the  triangle  shown  in 
Fig.  589  corresponding  to  the  line  2  1'  of  the  plan. 
Accordingly,  take  the  distance  1  2  of  the  plan  in  the 
dividers,  and  from  1  as  center  describe  a  short  arc. 
Then,  taking  the  distance  G  2  of  Fig.  589  in  the  di- 
viders, and  with  1'  as  a  center,  intersect  the  arc  already 
struck,  thus  establishing  the  point  2  of  the  pattern. 


Fig.  590.— The  Pattern  of  Article  Shown  in  Fig.  587. 


The  elements  of  the  second  triangle  are  the  distance 
1'  2'  on  the  inner  circle  of  the  plan,  the  hypothennse 
of  the  triangle  in  Fig.  588  corresponding  to  the  line 
2  2'  in  the  plan  and  the  side  just  established  in  the 
pattern  2  1'.  From  1'  as  center,  with  1'  2'  of  1he 
plan  as  radius,  describe  a  short  arc.  Then  from  2  as 
center,  with  E  2  of  Fig.  588  as  radius,  describe  a 
second  arc,  intersecting  the  one  already  made,  thus 
establishing  the  point  2'.  Proceed  in  this  manner 
until  triangles  have  been  described  adjacent  to  each 


Puttn'ii 


341 


other,  corresponding  to  the  divisions  first  established 
in  the  plan.  Then  linos  traced  through  the  points 
thus  established,  as  shown  from  1  to  10  and  from  1'  to 
In',  will  constitute  the  pattern  of  half  the  article. 
The  other  half  may  be  obtained  by  any'  convenient 
means  of  duplication  and  may  be  added  on  to  either 
end  of  the  half  already  obtained,  according  as  it  is 
desired  to  make  the  joint  at  the  widest  or  narrowest 
part  of  the  pattern. 


Fiij.  r,91.— Model  of  One-Half  the  Article  Shown  in   Fig.  587,  Illus- 
trating the  Construction  and  Use  of  the  Trinngles. 


In  Fig.  591  is  shown  a  model  which  may  be  con- 
structed of  thin  metal  and  wires,  or  of  cardboard  and 
threads,  according  to  convenience,  which  will  assist 
the  student  in  forming  a  correct  idea  of  the  relation- 
ship existing  between  the  various  lines  drawn  upon 
the  plan  and  the  lines  of  which  the  pattern  is  con- 


structed. The  top  and  bottom  of  the  model  are  dupli- 
cates of  the  inner  and  outer  circles  of  the  plan.  The 
piece  forming  the  bottom  should  have  the  solid  lines 
and  the  inner  circle  of  the  plan  drawn  upon  it  as  a 
means  of  placing  in  position  the  several  triangular 
pieces  shown,  which  are  duplicates  of  the  several 
triangles  shown  in  Fig.  588.  These  triangular  pieces 
having  been  placed  in  position  according  to  their 
numbers,  and  fastened  at  the  top  and  bottom,  their 
outer  edges,  or  hypothenuses,  will  then  represent  the 
solid  lines  drawn  across  the  pattern,  and  will  bear  the 
same  relation  or  angle  to  the  edges  of  the  top  and 
bottom  pieces  of  the  model  that  the  solid  lines  of  the 
pattern  bear  to  the  top  and  bottom  outlines  of  the  pat- 
tern. Finally,  threads  or  wires  having  been  attached, 
as  shown,  will  represent  the  dotted  lines  drawn  across 
the  pattern,  and  will  bear  the  same  angle  to  the  edges 
of  the  solid  triangles,  as  measured  upon  the  model, 
that  the  dotted  lines  of  the  pattern  bear  to  the  solid 
lines,  as  measured  upon  the  pattern. 

.Since  the  top  and  the  bottom  of  the  shape  here 
shown  are  both  round  and  horizontal,  the  figure  be- 
comes that  of  the  frustum  of  a  scalene  cone;  and, 
therefore,  its  sides,  if  continued  upward,  would  ter- 
minate in  an  apex  which  can  be  made  the  common 
apex  of  a  number  of  triangles  whose  bases  are  the 
spaces  upon  the  outer  line  of  the  plan.  This  method 
of  solving  the  problem  as  applied  to  a  full  scalene 
cone  is  given  in  Problem  163,  which  see. 


PROBLEM   184. 


The  Pattern  for  a  Flaring:  Article,  Round  at  Top  and  Bottom   one  Side  Being  Vertical. 


In  Fig.  592,  A  B  C  D  shows  the  elevation  of  the 
article,  below  which  E  F  G  H  shows  the  plan  at  the 
bottom  and  E  J  K  L  the  plan  of  its  top,  both  circles 
being  tangent  at  the  point  E. 

Divide  the  circle  representing  the  plan  of  the  top 
into  any  convenient  number  of  equal  spaces,  as -repre- 
sented by  the  small  figures  between  K  L  E  in  the  dia- 
gram. In  the  illustration  only  one-half  of  the  plan  has 
been  divided,  which  is  sufficient  for  the  purpose.  Next 
divide  a  like  portion  of  the  plan  of  the  base  into  the 
same  number  of  equal  parts,  as  shown  by  the  figures 


between  E  H  G.  Connect  these  two  sets  of  points, 
first  by  lines  drawn  between  like  numbers,  as  1  and  1', 
2  and  2',  3  and  3',  etc.  In  a  like  manner  connect  1 
of  the  inner  circle  with  2'  of  the  base,  2  with  3',  3  with 
4',  etc.,  all  as  shown  by  the  dotted  lines  in  the  plan. 
These  lines  just  drawn  are  the  bases  of  a  number  of 
right-angled  triangles,  whose  altitudes  are  equal  to  the 
vertical  hight  of  the  article,  and  whose  hypothenuses, 
when  obtained,  will  give  the  correct  measurements 
across  the  pattern  between  the  numbered  points. 

For  a  diagram  of  triangles  representing  the  solid 


342 


77te  New  Metal   Worker  Pattern  Book. 


linos  in  plan  erect  the  vertical  line  P  S  in  Fig.  593, 
equal  to  A  B  of  elevation.  Then  at  right  angles  from 
S  lay  off.  a  base  line,  upon  which  set  off  distances, 
measuring  from  S,  equal  to  the  lengths  of  the  several 
solid  lines  drawn  across  the  plan  in  Fig.  592.  Thus 
make  S  R  equal  to  K  Gr  (1  1')  and  S  2  to  2  2',  and  so 


Fig.  592.— Plan  and  Elevation  of  Flaring  Article,  Showing  Method  of 
Triangulation. 


on.  From  the  points  thus  established  in  the  base  draw 
lines  to  the  apex.  Then  the  hypothenuses  of  the  tri- 
angles will  be  equal  to  measurements  on  the  surface  of 
the  finished  article  on  lines  drawn  from  the  points  in  the 
base  to  corresponding  points  in  the  top.  In  the  same 


way  construct  the  diagram  representing  the  triangles 
Imsed  on  the  dotted  lines  in  plan,  as  shown  in  Fig.  ;V,»4. 
Set  off  T  V  equal  to  the  straight  higlit  of  the  article. 
From  V  draw  the  horizontal  line  V  U,  upon  which, 


7          »          54321 


Fig.  593. — Diagram  of  Triangles  Based  upon  Solid  Lines  in  the  Plan 
in  Fig.  592. 


measuring  from  V,  set  off  distances  equal  to  the  length 
of  the  dotted  lines  across  the  plan.  Thus  make  V  2 
equal  to  1  2',  V  3  equal  to  2  3',  etc.  From  the  points 
thus  established  in  V  U  draw  lines  to  the  apex  T. 
These  lines  will  be  equal  to  measurements  upon  the 


\Vx\x 


\  \\ 

X    X\N 

\     \\\\ 


10     118  S          7          6       5      4     S  8 

Fig.  594. — Diagram  of  Triangles  Based  Upon  Dotted  Lines  in  the 
Plan  in  Fig.  592. 


surface  of  the  finished  article  between  the  points  con- 
nected  by  the  dotted  lines  in  the  plan. 

Having  obtained  the  correct  dimensions  of  all  the 


Pattern  Problems. 


343 


triangles  assumed  at  the  beginning  of  the  work  they 
may  now  be  constructed  consecutively,  thus  developing 


8' 


Fig.  595. — The  Pattern  of  Flaring  Article  Shown  in  Fig.  592. 

the  pattern  in  the  following  manner:     Assume  any 
straight  line,  as  D  C  of  Fig.  595,  which  make  equal  to 


D  C  of  the  elevation,  or,  what  is  the  same  thing,  P  R  o* 
the  diagram  of  triangles,  Fig.  593.  From  C  as  a  cen- 
ter, with  a  radius  equal  to  1'  2'  of  the 
plan,  strike  a  small  arc,  which  inter- 
sect with  another  small  arc  struck 
from  D  as  center,  with  a  radius  equal 
to  T  2  of  Fig.  594,  thus  establish- 
ing the  point  2'  of  the  pattern. 
From  D  as  center,  with  a  radius 
equal  to  1  2  of  the  plan  of  the  top, 
Fig.  592,  strike  a  small  arc  and  inter- 
sect it  with  another  struck  from  2' 
of  the  pattern  as  center,  and  P  2  of 
Fig.  593  as  a  radius,  thus  establish- 
ing the  point  2  in  the  top  of  the 
pattern.  Proceed  in  this  manner, 
using  the  hypothenuses  of  the  tri- 
angles in  Fig.  594  with  the  spaces 
in  the  Outer  curve  of  the  plan,  Fig. 
592,  to  establish  the  points  in  the 
bottom  curve  of  the  pattern;  and 
the  hypothenuses  of  the  triangles  in 
Fig.  593  with  the  spaces  in  the  inner 
curve  of  the  plan  to  establish  the 
points  in  the  top  curve  of  the  pat- 
tern. Lines  traced  through  the 
points  of  intersection,  as  shown  from 
C  to  B  and  from  J)  to  A,  will,  with 
D  C  and  A  B,  constitute  the  pattern 
for  the  half  of  the  article  shown  by 
E  H  G  K  of  the  plan.  The  other 
half  may  be  added,  as  shown,  by 
any  convenient  means  of  duplica- 
tion. 

Since  the  top  and  the  bottom 
of  this  figure  are  both  round  and 
horizontal,  it  becomes  the  frustum 
of  a  scalene  cone,  which  permits  of 
its  being  treated  by  a  different,  and 
perhaps  simpler,  method  of  triangu- 


lation,  all  of  which  is  given  in  Problem  167,  to  which 
the  reader  is  referred. 


PROBLEM  185. 

-  The  Pattern  of  an  Article  having:  an  Elliptical  Base  and  a  Round  Top. 


Fig.  596  shows  the  plan  and  elevation  of  the  ar- 
ticle for  which  the  pattern  is  required.  Divide  one- 
quarter  part  of  the  plan  of  the  base  E  G  into  any 


convenient  number  of  equal  spaces,  and  divide  a  cor- 
responding part  of  the  plan  of  the  top  L  K  into  the 
same  number  of  spaces,  numbering  the  points  of  divi- 


344 


The  Neiv  Metal    Worker    l>,itl>-ni   Rook. 


sion  the  same  in  both,  as  indicated  by  the  small  figures 
lr  2,  3  and  I1,  2',  3',  etc.  The  article  here  shown  pos- 
sesses some  of  the  general  features  of  the  cone  in  that 
it  is  tapering  in  its  sides,  but  inspection  will  show  that 
the  slant  or  taper  of  its  sides  varies  in  different  pails 
of  its  circumference,  or  in  other  words,  that  different 


Fig.  596.— Elevation  and  Plan  of  Flaring  Article  with  Elliptical 
Base  and  Round  Top. 

lines  drawn  through  like  numbers  in  the  base  and  top 
would,  if  extended  upward,  meet  the  axis  at  different 
hights,  hence  some  means  must  be  devised  for  meas- 
uring the  real  distances  between  the  points  in  the  base 
and  the  points  in  the  top,  which  may  be  accomplished 
in  the  following  manner:  First  connect  all  points  in 
the  base  in  plan  with  points  of  the  same  number  in 

c-  C2 


/ 


<7 

o////// 


23  i  507 


Fig.  597.— Diagram  of  Triangles 
Rased  upon  Solid  Lines  of  the 
Plan  in  Fig.  596. 


Fig.  598. — Diagram  of  Triangles 
Based  upon  Dotted  Lines  of 
the  Plan  in  Fig.  596. 


the  top  by  means  of  a  solid  line,  as  shown  upon  the 
plan  by  lines  1,  I1,  2,  2',  etc.  Also  draw  the  inter- 
mediate dotted  lines  connecting  alternate  points,  as 
shown  in  the  engraving  by  2  1',  3  2'  4  3',  etc.,  thus 
dividing  the  entire  surface  of  the  article  into  triangles. 
Construct  a  diagram,  as  shown  by  A1  N1  C1,  Fig. 
597,  in  which  the  actual  distance  between  correspond- 
ing points  in  base  and  top  shall  be  shown.  Make 
C1  N1  equalto  the  straight  hight  of  the  article,  C  N  of 


the  elevation.  At  right  angles  to  it  set  off  N1  A1,  in 
length  equal  to  the  distance  1'  1  in  plan.  From  N' 
set  off  also  on  N'  A'  spaces  corresponding  to  2'  2, 
3'  3,  41  4,  etc.,  of  the  plan,  and  from  each  of  these 
points  draw  a  line  to  C1,  as  shown.  Then  the  lines 
converging  at  C1  represent  the  distances  which  would 
be  obtained  by  measurements  made  at  corresponding 
points  upon  the  article  itself.  Construct  a  like  dia- 
gram of  the  distances  represented  in  the  dotted  lines 
in  the  plan,  as  shown  by  C3  N2  O,  Fig.  598.  Make 
C'  N"  equal  to  C  N  of  the  elevation,  and  from  Na  set 
off  at  right  angles  the  line  N"  0.  Upon  this  line  make. 
the  spaces  N"  2,  N"  3,  N"  4,  etc.,  equal  to  the  length 
of  the  dotted  lines  1'  2,  2'  3,  3'  4,  etc.,  and  from  the 


Fig.  599.— The  Pattern  of  Article  Shown  in   l-'iij.  r,M. 

points  thus  obtained  in  W  O  draw  lines  to  C".  Then 
these  converging  lines  represent  the  same  distances  as 
would  be  obtained  if  measurements  were  made  between 
corresponding  points  upon  the  completed  article. 

For  the  pattern,  commence  by  drawing  any  line,  as 
P  X  in  Fig.  599,  on  which  set  off  a  distance  equal  t<> 
C'  1  of  the  first  diagram,  as  shown  by  1  I'.  Then, 
with  the  distance  from  1  to  2<>f  the  plan  for  radius,  and 

I  in  pattern  as  center,  describe  an  arc,  which  intersect 
by  another  arc  struck  from  I1  of  the  pattern  as  center, 
and  C"  2  of  the  second  diagram   as  radius,  thus   estab- 
lishing the  point  marked  2  in  the  pattern.      Next,  with 

II  2'  of  the  plan  as  radius,  and  from  1'   of   the   pattern 
as  center,  describe  an  arc,  which   intersect  bv  another 


Pattern    Problems. 


345 


arc  drawn  from  2  as  center,  and  with  C'  2  of  the  first 
diagram  as  radius,  thus  locating  the  point  2' of  the  pat- 
tern. Continue  in  this  manner,  locating  eacli  of  the 
several  points  shown  from  X  to  Y  and  from  P  to  R  of 
the  pattern,  through  the  several  intersections,  tracing 


the  lines  of  the  pattern,  as  shown.  Then  X  Y  R  P 
will  lie  one-quarter  of  the  required  pattern.  Repeat 
this  piece  three  times  additional,  as  shown  by  W  X  P 
T,  V  WTU  and  Y  Z  S  R,  reversing  each  alternate 
piece,  thus  completing  the  pattern. 


PROBLEM  186. 

Pattern  for  an  Irregular  Flaring  Article  which  is  Elliptical  at  the  Base  and  Round  at  the  Top,  the  Top 
being  so  Situated  as  to  be  Tangent  to  One  End  of  the  Base  when  Viewed  in  Plan. 


In  Fijr.  'il>()-  lct  1}  <">  ^  K  he  the  side  elevation  of 
the  article  and  K  N  M  one-half  of  the  plan  of  the  base. 
The  half  plan  of  the  top  is  shown  by  K  W  L,  the  base 
and  top  being  tangent  in  plan  at  the  point  K.  The 
conditions  and  method  of  procedure  in  this  problem 
do  not  differ  materially  from  those  of  Problem  184. 


Next  construct  the  diagrams  of  triangles,  as  shown 
in  Fig.  601  at  the  right  of  the  elevation,  making  A  U 
in  hight  equal  to  D  E  of  the  elevation,  and  lay  off  U  T 
at  right  angles  to  it.  Let  A  represent  all  points  in  the 
circle  representing  the  plan  of  the  top  of  the  article. 
Lay  off  from  U  upon  U  T  the  distance  from  each  of 


Fig.  601.— Diagrams  of  Triangles  Obtained 
from  Fiij.  600. 


Fig.  600.— Elevation,  and  Plan  of  an  Irregular  Flaring  Article 
with  Elliptical  Base  and  Round  Top. 


The  surface  of  this  article  may  be  divided  into 
measurable  triangles  in  the  following  manner :  Divide 
the  plans  of  ton  and  base  into  the  same  number  of 
equal  parts,  as  shown  by  1,  2,  3,  etc.,  in  the  base  and 
1',  21,  3',  etc.,  in  the  top,  and  connect  similar  points  in 
the  two  bv  solid  lines,  as  shown  by  6  6',  5  5',  etc.  Also 
connect  each  point  in  the  plan  of  the  top  with  the  next 
lower  number  in  the  plan  of  the  base,  as  shown  by  the 
dotted  lines  in  the  engraving,  as  6  7',  5  6',  etc. 


the  several  points  in  the  circle  to  the  corresponding 
point  in  the  ellipse.  Thus  make  U  7  equal  to  7'  7  of 
the  plan,  U  6  equal  to  61  6,  etc. ,  and  draw  the  radial  lines 
A  2,  A  3,  etc.  In  like  manner  construct  a  correspond- 
ing diagram,  as  shown  by  C  B  V,  using  for  the  spaces 
in  B  V  the  lengths  of  the  dotted  lines  between  the  circle 
and  the  ellipse  in  the  plan,  and  draw  C  2,  C  3,  C  4,  etc. 
By  means  of  these  two  sets  of  lines,  converging  at  A 
and  C  respectively,  and  the  stretchouts  of  the  two 


346 


The  New  Metal    Worker  Pattern  Book. 


curves  of  the  plan,  the  actual  dimensions  of  the  tri- 
angles into  which  the  surface  of  the  article  has  been 
divided  can  be  accurately  measured. 

These  are  to  be  used  in  describing  the  pattern  as 
follows :  At  any  convenient  place  draw  the  straight 
line  P  R  in  Fig.  602,  in  length  equal  to  G  F  of  the 
elevation,  or,  what  is  the  same,  equal  to  A  7  of  the 
first  diagram.  As  but  half  of  the  plan  of  the  article 
is  shown,  the  pattern  will  also  appear  as  one-half  of 
the  whole  shape,  and  therefore  P  R  will  form  its  cen- 
tral line.  From  P  as  center,  with  radius  C  6  of  the 
second  diagram,  describe  an  arc,  which  intersect  by  a 
second  arc  struck  from  R  as  center,  with  radius  7  6  of 
plan,  thus  establishing  the  point  6  of  the  pattern. 
Then  with  radius  A  6  of  the  first  diagram,  from  6  of 
the  pattern  as  center,  describe  an  arc,  which  cut  with 
another  arc  struck  from  7'  of  the  pattern  as  center,  and 
7'  6'  of  the  plan  as  radius,  thus  locating  the  point  6'  of 
the  pattern.  Continue  this  process,  locating  in  •  turn 
5,  5',  4,  41,  etc.,  until  points  corresponding  to  all  the 
points  laid  off  in  the  plan  are  established.  Lines  traced 


Fig.  60S. — Pattern  for  Article  shown  in   Fig.  600, 

through  the  points  7,  6,  5,  etc.,  and  7',  6',  5',  etc., 
will,  with  P  R  and  0  S,  form  the  patter*1  of  one-half 
the  article. 


PROBLEM    187. 

Pattern  for  a  Flaring  Article  or  Transition  Piece  Round  at  the  Top  and  Oblong:  at  the  Bottom,  the 

Two  Ends  Being  Concentric  in  Plan. 


ELEVATION 

C 


END  ELEVATION 


Q, 

P 
PLAN 

Fig.  60S. — Plan  and  Elevations  of  Flaring  Article  Bound  at  the  Top 
and  Oblong  at  the  Bottom. 


In  Fig.  603  are  shown  the  side 
and  end  elevations  and  the  plan  of  an 
article    which   might  form   a  transi- 
tion between  an  oblong  pipe  below 
and  a  round  pipe  above.    According 
to  the  conditions,  as  given  in  the  en- 
graving,  the  problem  is  capable  <>f 
two  solutions.      Since  the  upper  and 
lower    buses     ;irc    composed,    either 
wholly    or    in    part,   of    semicircles 
lying   in   parallel  planes,  those   por- 
tions  of  the  pattern   of  the  article  lying  between    the 
semicircles,  as  P  0  N  OP,  must  necessarilv  form  parts 
of  the  envelope  of  a  scalene  cone.     Those  portions  of 
the  pattern    ma}-  therefore  be    obtained,  if  desirable, 
by  the  method  employed  in  Problems   168  and  169. 
The  other  solution,   which  is  perhaps  the  nn in- 
simple,  is  given  in   Figs.  ti<)4  to  606.      An  inspection 
of  the  plan  will  show  that  the  article  consists  of  l'< un- 
like quarters,    therefore  in  Fig.  ti<>4   is  shown  an  en- 
larged plan  and  elevation  of  one-quarter  of  the  article. 
Divide    P  T    and    0  N    of    the    plan    eaeli    into  the 


Pattern   Problems. 


347 


same  number  of  equal  parts,  and  connect  the  points 
in  P  T  with  those  in  0  N.  as  indicated  by  the  solid 
lines.  Also  connect  points  in  the  top  with  those  in 


of  dotted  lines  in  plan.  Thus  make  W  7  equal  to  T  7 
and  W  8  equal  to  2  8,  and  so  on.  From  the  points 
thus  established  in  the  base  draw  lines  to  V.  The 


ELEVATION 


K      H 


6  8910  |_ 


W 


9,o8 


PLAN 


-UN 


Fiij.  604.- 


-Quarter  Plan  and  Elevation  Enlarged,  Showing  Method 
of  Triangulation. 


the  bottom,  as  shown  by  the  dotted  lines.  These  lines 
represent  the  bases  of  right  angled  triangles,  the  alti- 
tude of  which  will  be  equal  to  the  straight  hight  of  the 
article.  For  a  diagram  of  the  triangles  representing  the 
solid  lines  of  the  plan,  draw  any  vertical  line,  as  J  K  in 
Fig.  605,  which  make  equal  in  hight  to  the  hight  of  the 
article,  as  shown  by  the  dottedjine  D  F.  From  K,  at 
right  angles  to  J  K,  draw  K  L,  upon  which  set  off  dis- 
tances, measuring  from  K,  equal  to  the  lengths  of  the 
solid  lines  drawn  across  the  plan.  Thus  make  K  6 
equal  to  T  N  and  K  7  equal  to  2  7  of  plan,  and  so  on. 
Also  set  off  from  K  the  distance  II  P  of  plan,  as 
shown  by  K  H.  From  the  points  thus  established  in 
K  L  draw  lines  to  J.  The  hypothenuses  thus  obtained 
will  give  the  distances  across  the  finished  article,  as  in- 
dicated by  the  solid  lines  of  the  plan. 

The  next  step  will  be  to  construct  a  diagram  of 
triangles  that  will  give  the  distances  between  points  in 
tin1  base  and  top,  as  indicated  by  the  dotted  lines  in 
plan.  This  diagram  is  constructed  in  a  similar  man- 
ner, as  shown  at  the  right  in  Fig.  605.  Draw  the  right 
angle  V  W  X,  making  V  W  equal  to  tl.e  straight  hight 
of  the  article,  and  from  W  set  off  on  W  X  the  lengths 


A'lV/.  G05.—  Diagram  of  Triangles. 


livpothenuses  of  the  triangles  thus  obtained  will  give 
the  distance  from  points  in  the  base  to  points  in  the 
top,  as  indicated  by  the  dotted  lines  in  plan. 

To  lay  out  the  pattern  first  draw  any  line,  as  T  N 
of  pattern,  in  length  equal  to  J  6  of  first  diagram  of 
triangles,  or,  which  is  the  same  thing,  D  E  of 
elevation. '  From  N  of  pattern  strike  a  short  arc 


Firj.  606.— Quarter  Pattern  of  Article  Shown  in  Fig.  60S. 

with  a  radius  equal  to  N  7  of  the  plan,  as  shown. 
From  T  of  pattern  as  center,  with  radius  equal  to  V  7 
of  the  second  set  of  triangles,  intersect  this  arc,  thus 
establishing  the  point  7  of  pattern.  From  T,  with 
radius  equal  to  T  2  of  the  plan,  strike  a  small  arc,  as 
shown,  and  intersect  it  with  another  from  point  7  of 
pattern  as  center,  with  J  7  of  the  diagram  of  triangles 
as  radius,  thus  establishing  the  point  2  in  pattern. 


348 


Tfie  New  Metal    Worker  Pattern  Book. 


Proceed  in  this  manner,  using  alternately  the  hypoth- 
enuses  of  the  triangles  in  V  W  X  of  Fig.  605,  the 
spaces  in  plan  of  base  0  N,  the  hypothenuses  of  the 
triangles  in  J  K  L,  Fig.  605,  and  the  spaces  in  the  plan 
of  top,  P  T,  in  the  order  named,  and  as  above  ex- 
plained. The  resulting  points,  as  indicated  by  the 
small  figures  in  the  pattern,  will  be  points  through 


which  the  pattern  line  will  pass.  For  the  pattern  of 
triangle  P  H  O  of  pattern,  with  0  of  pattern  as  center, 
and  O  II  of  plan  as  radius,  strike  a  small  arc  in  the 
direction  of  II.  With  P  of  pattern  as  center,  and  .1  IL 
of  the  diagram  of  triangles  as  radius,  describe  another 
small  arc  intersecting  the  one  just  struck.  Draw  ()  11 
and  II  P,  thus  completing  the  quarter  pattern. 


PROBLEM   188. 

Pattern  for  a  Transition  Piece  Round  at  the  Top  and  Oblong  at  the  Bottom.    Two  Cases. 


In  Fig.  607  is  shown  the  plan  and  elevations 
of  a  transition  piece,  constituting  the  first  c«w,  such 
as  is  frequently  required  in  furnace  work  when  it 
is  necessary  to  connect  a  round  pipe  with  another 
pipe  of  equal  area  but  flattened  into  an  oblong 
shape. 

In  Fig.  611  are  shown  the  plan  and  elevations  of  a 


SIDE  VIEW. 


98  7    65432 


101112  13 


first  case,  shown  in  Fig.  6ii7.  The  principle  involved 
in  developing  the  patterns  of  the  two  shapes  is  exactly 
the  same,  consequently  the  following  demonstration 
will  apply  equally  well  to  either  Fig.  607  or  611,  in 
each  of  which  corresponding  points  are  lettered  the 
same.  (It  will  be  noticed  that  separate  diagrams  of 
triangles  and  a  separate  pattern  corresponding  to  Fig. 


Fig.  608. — Diagrams  of  Triangles  Based  upon  the  Solid  Lines  of  L 

the  Plan,  Fig.  607.  Fig-  607.— Plan  and  Elevation  of  Transition  Piece— First  Case. 


transition  piece  of  the  second  cose,  answering  the  same 
general  description  as  that  given  above,  but  differing 
only  in  the  fact  that  the  circle  representing  the  top  in 
the  plan  view  does  not  touch  the  side  of  the  line  rep- 
resenting the  bottom  of  the  article.  In  other  words, 
the  side  F  H  is  slanting  instead  of  vertical,  as  in  the 


fill  have  not  been  given.  While  in  reality  they  would 
differ  somewhat  from  those  shown  in  Figs.  608  to  Uln. 
they  would  have  the  same  general  appearance,  and  in 
method  of  construction  would  be  exactly  the  same,  and 
therefore  have  not  been  considered  necessary  to  the 
study  of  the  problem.) 


349 


"N  P  0  Q  represents  a  plan  (if  the  shape  described, 
above  which  A  B  1)  C  shows  an  elevation  of  the  front 
of  the  same,  or  as  seen  when  looking  toward  Q.  while 
to  the  left  is  shown  a  view  obtained  by  looking  to- 
ward the  side  X,  in  which  K  (1  corresponds  to  P  X  of 
the  plan  and  F  II  to  Q'  g. 

Divide  one-half  of  the  plan  of  the  top  or  round 
portion  of  the  article  into  anv  convenient  number  of 
equal  spaces,  in  this  case  l:>.  Since  by  the  conditions 
of  the  problem  one-half  of  the  round  end  corresponds 
to  the  semicircular  end  of  the  oblong  part,  divide  the 
semicircle  J  N  L  into  the  same  number  of  equal  parts. 


V3 

V'  , 


654     31     1 


7  8    9    10  II   12 


Fill.  009. — Diar/rams  of  Triangles  Based  upon  the  Dotted  Lines 
of  the  Plan,  Fig.  607. 


Then  connect  points  in  the  two  lines  of  the  plan  of  the 
same  numerals.      For  example,  1  with  1,    2  with   '2,    :•> 
with  .'!,  etc.     In  like  manner  connect  the 
points  in  the  end  of  the  oblong   portion 
with    points    of    the    next    higher   num- 
ber   in    the    round    end,  as    shown,   as, 
for  example,  1  with  2,  2  with  3,  3  with 
4,  etc.      Upon    all  of  these   lines  drawn 
in     the    plan     it    will    be    necessary  to 
construct  sections  or  triangles   in  which 
these     lines     form    the     bases    and    in 
which  the   vertical   hight  of  the  article 
\V    V    is    the    altitude.       The    various 
li ypothenuses    thus    obtained  will  then 
represent  the  true   distances  across  the 
linished  article  upon  the  lines  indicated 
in  the  plan.      The   triangles  correspond- 
in^  to  the  solid  lincsof  tlu;  plan  are  shown  in  Fig.  f>OS, 
while  those  corresponding  to  the  dotted  lines  of  the  plan 
are  shown  in  Fig.  <>u',».    In  order  to  avoid  confusion  each 
of  these  sets  has  been  divided  into  two  groups,  as  shown, 
and  are  constructed  as  follows:      Lay  off  at  any    con- 
venient   place    the  line  A    15  (Fig.   tlUS).  equal   to  V  W 
of  Fig.  t'.uT.     From  the  point  B,  and  at  right  angles  to 
A  B,  draw  the  line  B  C  and  upon  it  set  off  the  lengths 


of  the  several  solid  lines  connecting  the  two  outlines 
in  the  plan.  Thus  make  B  1  equal  to  the  distance  1  1 
or  J  P  of  the  plan.  B  2  is  equal  to  the  distance  2  2 
of  the  plan,  and  B  3  is  equal  to  the  distance  3  3  of  the 
plan.  etc.  As  already  explained,  D  E  is  a  duplicate 
of  A  B,  and  E  F  is  drawn  at  right  angles.  On  E  F 
the  spaces  K  10,  E  11,  E  12  and  E  13  are  set  off, 
being  equal  respectively  to  10  10,  11  11,  etc.,  of  the 
plan.  From  the  points  thus  established  in  the  base 
lines  B  (J  and  E  F  draw  lines  to  the  apices  A  and  D, 
thus  completing  the  triangles.  Then  the  hypothenuses 
A  1,  A  2,  A  3,  etc.,  D  13,  D  12,  etc.,. correspond  to 
the  width  of  the  pattern  measured  between  points  in- 
dicated by  like  figures  in  the  plan. 

In  the  same  general  manner  construct  the  triangles 
shown  in  Fig.  009,  which  correspond  to  sections  on 
the  dotted  lines  across  the  plan.  G  H  and  L  M  of  Fig. 
609  correspond  to  the  hight  V  "W  of  the  elevation. 
II  K  and  M  N  are  drawn  at  right  angles  to  the  perpen- 
diculars, and  on  these  base  lines  spaces  are  set  off, 
measuring  from  H  and  M  respectively,  corresponding 
to  the  length  of  the  dotted  lines  across  the  plan.  Thus 
H  1  corresponds  to  1  2  of  the  plan,  and  M.  12  corres- 
ponds to  12  13  of  the  plan.  From  the  points  thus  estab- 
lished in  the  base  line  lines  are  drawn  to  the  apices  G 
•and  L,  thus  completing  the  triangles.  These  hypothe- 
nuses are  equal  to  the  width  of  the  pattern  measured 


M      I?       13 
Fitj.  610.— Pattern  of  Transition  Piece  Shown  in  Fig.  607. 

between  points  connected  by  the  dotted  lines  in  the 
plan.  By  the  conditions  of  the  problem,  inasmuch  as 
there  are  straight  portions  in  the  oblong  end,  there  will 
bo  portions  of  the  pattern  that  will  correspond  to  tri- 
angles the  bases  of  which  are  equivalent  to  the  length 
of  the  straight  portion  in  the  plan  and  the  hights  of 
which  are  equal  respectively  to  the  distances  E  G  and 
F  II  of  the  side  view. 


350 


Tlie  New  Metal   Worker  Pattern  Book. 


SIDE  VIEW 


Therefore,  to  describe  the  pattern  proceed  as  fol- 
lows :  At  any  convenient  place,  as  shown  by  A  B  in 
Fig.  610,  draw  a  line  equal  to  the  width  of  the  pattern 
at  a  point  corresponding  to  Q1  Q  in  the  plan.  This 
would  be  the  same 
as  F  H  of  Fig.  607 
or  611.  To  com- 
plete the  triangu- 
lar portion  referred 
to  set  off  from  B 
the  distance  B  C 
equivalent  to  Q  L 
of  Fig.  60Y,  thus 
obtaining  the  point 
C.  The  dotted  line 
A  C  in  the  pat- 
tern is  drawn  to  show  the  portion  obtained  by  this 
means.  From  C  as  center,  with  the  space  13  12  of  the 
plan  of  the  oblong  end  as  radius,  describe  a  small  arc,  as 
shown  to  the  left.  Then  from  A  as  center,  with  radius 
L  12  of  Fig.  609,  corresponding  to  the  width  of  the  pat- 
tern measured  on  the  dotted  line  13  12  of  the  plan,  de- 
rscribe  another  arc  intersecting  the  one  just  drawn,  thus 
establishing  the  point  12  in  the  lower  edge  of  the  pat- 
tern. From  12  as  center,  with  D  12  of  Fig.  608  as 
radius,  being  the  width  of  the  pattern  on  the  line  12  12 
of  the  plan,  describe  a  short  arc,  as  shown  at  12  in  the 
upper  line  in  the  pattern.  Intersect  this  with  another 
arc  drawn  from  A  as  center,  with  13  12  of  the  plan  of 
the  round  end  as  radius,  thus  locating  the  point  12  in 
the  upper  line  of  the  plan.  Proceed  in  this  manner, 
using  in  the  order  described  the  stretchout  of  the  semi- 
circular end  of  the  oblong  section,  the  hypothenuses  of 
the  triangles  corresponding  to  the  dotted  .lines  in  the 
plan,  the  hypothenuses  of  the  triangles  corresponding  to 
the  solid  lines  in  the  plan  and  the  stretchout  of  the 
circular  end,  reaching  finally  the  points  D  and  E  of  the 
pattern,  representing  one  side  of  the  remaining  triangu- 


lar section  to  be  added.  From  E  as  center,  with  K  J 
of  the  plan  as  radius,  describe  an  arc.  From  D  as 
center,  with  radius  equal  to  D  E  of  the  pattern,  strike 
a  second  arc  intersecting  the  one  just  drawn,  thus 


Fig.  611.— Plan  and  Elevation  of  Transition  Piece— Second  Case. 

locating  the  point  F.  Connect  F  and  E  and  also  F  and 
D,  thus  completing  the  pattern  of  the  part  correspond- 
ing to  K  J  P  of  the  plan.  The  dotted  line  DG  drawn 
across  the  pattern  corresponds  to  the  line  X  P  of  the 
plan,  and  D  A  B  Gr  will  be  one  half  of  the  finished 
pattern. 


PROBLEM  189. 

Pattern  for  an  Offset  Between  Two  Pipes,  Oblong  in  Section,  whose  Long  Diameters  Lie  at  Right 

Angles  to  Each  Other. 


In  Fig.  612  are  shown  the  plan  and  elevations  of 
an  offset  or  transition  piece  to  form  a  connection  be- 
tween two  pipes  of  oblong  profile  which  will  be  spoken 
of  in  the  demonstration  as  the  »/</./<:/•  and  the  l<n/-<-r 
pipes.  M  N  0  P  Q  L  of  the  plan  is  the  section  or 


profile  of  the  upper  pipe  which  begins  at  the  line  A  B 
of  the  elevation  and  extends  upward,  while  F  i  II  I 
J  K  of  the  plan  is  the  section  of  lower  pipt'  which  he- 
gins  at  the  line  D  C  of  the  elevation  and  extends 
downward,  A  B  C  D  being  one  view  of  the  offset.  At 


Pattern  Problems. 


the  right  the  .same  plan  is  shown,  turned  one-quarter 
way  around,  from  which,  and  the  front  elevation,  are 
projected  a  side  elevation  of  the  offset.  Corresponding 
points  in  the  two  plans  arc  indicated  by  the  same  let- 
ters, capitals  bom;/  used  in  the  one  and  italics  in  the 
other. 

An  inspection  of  the  plan  will  show  that  the  long 
diameter  0  L  of  the  upper  pipe  cuts  the  profile  of  the 
lower  pipe  nearer  one  end  than  the  other,  from  which 


the  offset.  Lines  from  G  H  of  the  bottom,  to  N  0  of 
the  top  would  form  another  corner,  lines  from  I  J  up 
to  O  P  a  third  corner  and  lines  from  J  K  up  to  Q  L 
the  fourth.  Lying  between  these  corner  pieces  are 
the  two  triangular  end  pieces  K  L  F  and  I  II  0  and 
the  side  pieces  M  N  G  and  Q  P  J.  For  convenience 
in  describing  the  pattern  the  joint  will  be  assumed 
through  one  of  the  ends  at  the  line  E'  L. 

Preparatory  to  obtaining  the  pattern,  first  divide 


o 

PLAN 


Fig.    612.— Plan    and    Elevations   of  an    Offset  between   Two  Pipes 
Oblong  Profile. 


"f 


it  must  be  concluded  that  the  pattern  cannot  be  com- 
posed of  symmetrical  halves  or  quarters  and  that  there- 
fore the  entire  pattern  must  be  developed  at  one  oper- 
ation. 

As  the  sections  of  both  pipes  may  be  said  to  con- 
sist of  four  quarter  circles  joined  to  the  straight  side, 
the  pattern  will  consist  principally  of  four  rounded 
corners  joining  the  quarter  circles  which  occupy  the 
same  relative  position  in  the  two  pipes.  Thus  lines 
joining  the  quarter  circles  F  G  of  the  bottom  and  L  M 
of  the  top  would  form  one  corner  of  the  envelope  of 


each  of  the  four  quarter  circles  of  the  lower  pipe  into 
the  same  number  of  equal  parts ;  also  divide  the  pro- 
tile  of  the  upper  pipe  in  the  same  manner.  To  avoid 
confusion  of  lines  two  separate  plans  are  shown  in 
Figs.  613  and  614  for  obtaining  these  divisions.  In 
Fig.  613  are  shown  the  divisions  of  what  may  be  called 
the  back  end,  while  Fig.  614  shows  those  of  the  front 
end.  As  will  be  seen  by  these  plans,  the  points  have 
been  numbered  alternately  in  the  bottom  and  the  top. 
Solid  lines  are  first  drawn,  as  shown,  connecting  points 
1  2,  3  4,  5  6,  etc.,  after  which  the  four-sided  figures  thus 


352 


New  Metal    Worker  Pattern  Book. 


produced  are  divided  diagonally  by  the  dotted  lines 
1  4,  3  6,  5  8,  etc. 

The  next  operation  is  the  construction  of  a  series 


between  F  and  G  in  the  lower  pipe  to  points  betwr 
M  and  L  of  upper  pipe,  as  indicated  by  the  solid  HIK 
in  plan.     For  the  diagram  of  triangles  representing  ti; 


PLAN 


Duplicates  of  Plan  in  Fit).  611,  Showing  System  of  Triangulation. 


of  right  angled  triangles  whose  bases  shall  be  equal  to 
the  several  solid  and  dotted  lines  which  have  just 
been  drawn,  and  whose  altitudes  shall  equal  the  straight 
hight  of  the  offset.  These  have  been  arranged  in  four 
groups,  shown  in  Figs.  615  tp  618,  corresponding  to 
the  four  corner  pieces  above  mentioned.  In  Fig.  615, 
draw  the  right  angle  R  S  T,  making  R  S  equal  to  the 
straight  hight  of  the  article,  as  indicated  by  A  Z  of 
front  elevation.  Measuring  from  S,  set  off  on  S  T  the 
lengths  of  solid  lines  in  F  G  M  L  of  the  plan  Fig.  613, 
including  L  E,  which  will  give  the  slant  hight  cor- 


dotted  lines  in  F  G  M  L  of  the  plan,  draw  the  right 
angle  R'  S1  T',  as  shown  at  the  right  in  Fig.  615, 
making  R1  S1  equal  to  the  straight  hight  of  the  article, 
as  derived  from  A  Z  of  front  elevation.  Measuring  in 

R. 


Fig.  616.— Diagrams  of  Triangles  in  O  H  O  N  of  Fig.  614. 

responding  with  I  c  of  the  side  elevation.  Connect  the 
points  in  S  T  with  R,  as  shown  by  the  solid  lines. 
These  hypothenuses  represent  the  distances  from  points 


E102 


Fig.  615.—  Diagrams  of  Triangles  in  F  G  M  L  of  Fig.  613. 

each    instance   from    S1,   set  off  on 
S1  T'  the  lengths  of  dotted  lines  in 
F  G  M  L,  and  from  the  points  thus 
obtained   draw    lines    to    R1.      The 
diagrams    shown    in    Fig.    616    are 
constructed  in  the  same  manner  and 
correspond  to  the  solid   and  dottrel 
lines    in    the   corner  G  II  0  N,   shown  in   Fig.    614. 
The    diagrams  (if    triangles  in  Fig.    617    are    derived 
from  the  solid  and  dotted  lines  in   I  J  P  0  of  1(  ig. 


Pattern  Problems. 


353 


<>14,  and  in  Fig.  til 8  are  shown  the  diagrams  of  tri- 
angles derived  from  the  solid  and  dotted  lines  in  Q  L 
K  J  of  Fig.  613. 


X 


I  hi1  plan  as 
Proceed  in 
dicated  by 


/•'/(/.  un.—l>i<i<jrains  nf  Triangles  in  IJ PO  of  Fiij.  614. 

To  develop  the  pattern  draw  anv  line,  as  E  L  of 
Fig.  i!l!i,  in  lengtli  equal  to  R  E  of  Fig.  615,  which 
gives  the  actual  distunec  from  E  in  the  base  to  L  in  the 
top,  as  also  shown  by  I  c  of  side  elevation  and  indicated 
i:i  corresponding  plan  by  I  e.  With  E  of  pattern  as 
center,  and  E  F  of  plan  as  radius,  strike  a  small  arc, 
F,  •which  intersect  with  one  struck  from  L  of  pattern 
as  center,  and  It  T  of  Fig.  615  as  radius,  thus  estab- 
lishing point  F  of  pattern.  "With  point  F  of  pattern 
as  center,  and  R'  1  of  Fig.  615  as  radius,  strike  a 
small  arc.  4,  which  intersect  with  one  struck  from 
point  L  of  pattern  as  center,  and  L  4  of  Fig.  613  as 


radius,  thus  establishing  point  3  of  pattern. 
this  manner,  as  above  described,  and  as  in- 
the  solid  and  dotted  lines,  until  the  points  G 
and  M  of  pattern  are  located.  With  M 
of  pattern  as  center,  and  M  N  of  the 
plan  as  radius,  strike  a  small  arc,  N, 
which  intersect  with  one  struck  from 
G  of  pattern  as  center,  and  U  W  of 
Fig.  616  as  radius,  thus  establishing 
point  N  of  pattern.  Proceed  in  the 
manner  indicated  until  the  remaining 
points  in  the  pattern  are  located.  It 
will  be  observed  that  the  letters  and 
figures  in  pattern  designate  points 


x 


\ 


Y2 


32 
SO 


\ 


% 


Y3 


S3  27 
III 
31 


Fi(j.  618. — Diagrams  of  Triangles  in  (J  L  K  J  uf  Fig.  61S,     . 


619.—  Pattern  of  Offset  Shown  in  fig.  612. 


raiiius,  thus  establishing  point  4  of  pattern.  With 
pc'/.it  4  of  pattern  as  center,  and  R  4  of  Fig.  615  as 
radiu",  strike  a  small  arc,  3,  which  intersect  with  one 
struck  from  point  F  of  pattern  as  center,  and  F  3  of 


similarly  indicated  in  Figs.  613  and  614.  Lines  traced 
through  the  points  obtained  as  directed,  and  as 
shown  from  E  to  E'  and  L'  to  L,  will  produce  the  de- 
sired pattern. 


354 


Tlie   Sctc   Metal     \Yurkcr   Ptnil^ni  Book. 


PROBLEM    190. 

Pattern  for  an  Irregular  Flaring  Article  Whose  Top  is  a  Circle  and  Whose  Base  is  a  Quadrant. 


In  Fig.  620  G  E  F  shows  the  plan  of  the  article 
at  the  base,  L  J  K  the  plan  at  the  top  and  A  B  C  D 
an  elevation  of  one  side.  An  inspection  of  the  plan 
will  show  that  the  article  consists  of  two  symmetrical 
halves  when  divided  by  the  line  G  H,  and  that,  there- 


G  F 

Fig.  620.— Plan  and  Elevation  of  an  Article  Whose  Tup  is  a   Circle 
and  Whose  Base  is  a  Quadrant. 

fore,  the  triangulation  of  one-half  will  answer  for  the 
whole.  On  account  of  the  dissimilarity  between  the 
outlines  of  the  top  and  the  bottom  some  judgment  will 
be  required  in  adopting  a  good  division  of  the  surface 
into  triangles. 

As  the  point  L  of  the  plan  is  the  nearest  point  to 
the  adjacent  side  E  G,  it  must  be  chosen  as  the  vertex 
<>f  a  triangle  whose  base  is  E  G.  That  portion  of  the 
circle  of  the  top,  therefore,  between  L  and  its  corre- 
sponding point  K  in  the  other  half  of  the  article  must 
be  considered  as  the  base  of  an  oblique  cone  whose 
apex  is  at  G. 

It  is  always  advisable  in  the  division  of  a  surface 
into  triangles  that  the  solid  and  dotted  lines  crossing 
the  plan  should  intersect  the  outlines  of  the  top  and 
the  bottom  as  nearly  at  right  angles  as  possible. 


Therefore,  since  the  remainder  of  the  top  (Lto.J)  and 
E  H  of  the  base  are  the  bases  of  a  surface  which  must  be 
so  divided  as  to  best  serve  the  purposes  of  triangulation, 
it  is  advisable  to  divide  L  .1  into  inure  spaces  than 
E  H,  allowing  the  extra  spares  in  the  top  nearest  the 

point  L  to  form  the  bases  of  a 
number  <>I  converging  triangles, 
as  shown.  Thus  lirst  divide  E 
II  into  any  suitable  number  of 
spaces,  as  shown  by  the  small 
I  ig  u  res  5  to  0,  then  divide.  L 
,J  into  a  greater  number  of  equal 
spaces  than  E  II,  as  shown  bv 
the  small  figures  '•'>  to  l».  Con- 
nect points  of  like  number  in 
the  two  outlines  by  solid  lines, 
commencing  at  II  and  .1.  as 

shown  from  i)  9  to  5  ~),  drawing 

lines  also  from  4  and  3  of  the  top  to  5  (E)  of  the  bot- 
tom. Also  draw  the  dotted  lines  5  6,  6  7,  etc.,  and 
the  solid  lines  from  points  in  L  N  to  G.  These  solid 
and  dotted  lines  will  then  form  the  bases  of  a  series  of 
right  angled  triangles  whose  hypothenuscs  will  give 
the  real  distances  across  the  envelope  of  the  finished 
article. 

These  triangles  are  constructed,  as  shown  in  Fig. 
621  at  the  right  of  the  elevation,  in  the  following  man- 
ner. Extend  A  B  and  1)  C  of  the  elevation,  through 
which  draw  any  vertical  line,  as  ( j  K.  From  Q  on  Q  I* 
set  off  the  lengths  of  all  the  solid  lines  of  the  plan. 
Thus  make  Q  9  equal  to  9  9  or  J  II  of  the  plan,  Q  8 
equal  to  S  8  of  the  plan,  etc.,  and  from  the  points  thus 
established  draw  lines  to  E.  In  like  manner  draw  the 
vertical  line  T  V,  and  from  T  on  T  S  set  off  the 
lengths  of  the  dotted  lines  of  the  plan,  as  shown  l>y 
the  small  figures,  and  from  the  points  thus  obtained 
draw  lines  to  V,  as  shown.  The  small  figures  in  S  T 
correspond  with  the  figures  in  L  .1,  the  top  line  of  the 
plan. 

In  laying  out  the  pattern  shown  in  Fig.  (>±2  the 
joint  is  assumed  upon  the  line  J  H  of  the  plan.  The 
pattern  may  be  best  begun  by  first  laying  out  one  of 
the  large  triangles  forming  a  side  of  the  article,  as 
E  L  G  or  G  K  F  of  the  plan,  shown  also  by  1)  N  C  of 
the  elevation,  Fig.  620.  Draw  any  hori/ontal  line,  as 
E  G  of  Fig.  622,  equal  in  length  to  E  G  of  the  plan. 


Problems. 


355 


From  E  us  center,  with  radius  R ?,  of  Fig.  621,  de- 
scribe :i  small  arc  near  I.,  which  intersect  with  another 
arc  drawn  from  G  as  ccntci1,  with  a  radius  equal  to 
It  3'  of  Fig.  021.  thus  establishing  the  point  L  of  the 
pattern.  From  (i  of  the  pattern  as  center,  with  radii 


E  G 

Fig.  fcy.— Pattern  of  Article  Shmrn  in  Fig.  620. 

equal  to  It  '2  and  R  lof  Fig.  621,  describe  small  ares,  as 
shown  between  L  and  K  of  the  pattern.  Take  between 
the  points  of  the  dividers  a  space  equal  that  used  in 
dividing  the  arc  L  K  of  the  plan,  and  placing  one  foot 
of  the  dividers  at  L  of  the  pattern  step  from  arc  to 
are,  reaching  K,  as  shown,  and  through  the  points 
thus  obtained  draw  L  K  of  the  pattern;  also  draw 
K  G.  From  E  of  the  pattern  as  center  ,  with  radii  equal 
to  R  4  and  It  5  of  Fig.  021,  describe  small  arcs  to  the 


left  of  L.  With  the  dividers  set  to  the  space  used  in 
dividing  the  arc  L  J  of  the  plan,  place  one  foot  at 
L  (3)  and  step  first  to  arc  4,  then  to  arc  5,  thus  es- 
tablishing the  points  4  and  5. 

With  the  last  obtained  point,  5,  of  the  pattern  as 
center,    and  a  radius    equal    to    the 
dotted  line  V  5  of  Fig.  621,  describe 
a  small  arc  (6'),  which  intersect  with 
another  arc  struck  from  point  E  of 
pattern  as  center,  with  a  radius  equal 
to  5  6  of  the  base  line  E  H  of  the 
plan,  thus  establishing  the  point  6' 
of  the  pattern.      With  6'  of  the  pat- 
tern as  center,  and  a  radius  equal  to  R 
5  of  Fig.  621,  describe  a  small  arc 
(6),  which  intersect  with  another  arc 
struck  from  5  of  pattern  as  center, 
with  a   radius  equal   to   5  6  of   the 
top  line  L  J  of  the  plan,  thus  estab- 
lishing the  point  6  of   the  pattern. 
Proceed  in  this  manner  in  the  con- 
struction of  the  remaining  triangles 
of  the  pattern,  using  alternately  the 
lengths  of  the  dotted  and  the  solid 
hypothenuses    in    Fig.    621    corre- 
sponding to  the  dotted  and  the  solid  lines  crossing  the 
plan,  in  the  order  in  which  they  occur,  to  determine 
the  width   of  the  pattern ;   the  spaces   in  E  II  of  the 
plan  to  form    the   lower  line  E  II  of  the  pattern  and 
the  spaces  in  L  J  of  the  plan  to  form  the  upper  line 
of    the    pattern,   all   as  shown.      The  remaining  parts 
of    the    pattern    can  %  be   obtained   by  any   convenient 
means   of  duplication,   K  F  G    being   a   duplicate   of 
LEG  and  K  J1  H'  F  being  a  duplicate  of  L  J  H  E. 


PROBLEM   191. 


The  Patterns  for  a  Three-Piece  Elbow,  the  Middle  Piece  of  which  Tapers. 


In  Fig.  623,  let  A  B  D  F  II  G  E  C  be  the  side 
view  of  a  three-piece  elbow,  the  middle  piece  (C  D  F  E) 
of  which  is  made  tapering.  The  piece  C  D  F  E  may 
also  be  described  as  an  offset  between  two  round  pipes 
of  different  diameters.  A  half  profile  of  the  upper 
and  smaller  of  the  two  pipes  is  shown  by  a  rfi  b,  while 


g  n  h  shows  that  of  the  larger  pipe.  The  straight  por- 
tions A  B  D  C  and  E  F  H  G  are  in  all  respects  similar 
to  inanv  pieces  whose  patterns  have  already  been  de- 
scribed in  the  first  section  of  this  chapter  in  Problems 
38  to  45  inclusive.  It  will  therefore  be  unnecessary 
to  repeat  the  description  in  this  connection. 


356 


Tfie  New  Metal   Worker  Pattern  Boole. 


Since  an  oblique  section  through  a  cylinder  is  an 
ellipse,  an  inspection  of  the  drawing  will  show  that  the 
sections  C  D  and  E  F,  the  upper  and  lower  bases  of 
the  middle  piece,  must  be  elliptical.  The  first  opera- 
tion, therefore,  will  be  to  develop  the  ellipses,  wliidi 
may  be  done  in  the  following  manner :  Divide  the 


fig  623.— Elevation  of  a  Three-Piece  Elbow,  the  Middle  Piece  of 
which  Tapers. 

profile  ami  into  any  convenient  number  of  equal 
spaces,  as  shown  by  the  small  figures.  From  the  points 
thus  obtained  drop  lines  vertically  to  a  b  and  continue 
them  till  they  cut  the  line  C  D.  From  the  intersec- 
tions on  C  D  carry  lines  at  right  angles  to  the  same 
indefinitely,  as  shown.  Through  these  lines  draw  any 


line,  as  c  il,  parallel  to  C  D.  Upon  each  of  the  lines 
drawn  from  C  D,  and  measuring  from  c  d,  set  off  the 
lengths  of  lines  of  corresponding  number  in  the  profile 
a  m  l>,  measuring  from  a  I.  A  line  traced  through  the 
points  thus  obtained,  as  shown  by  c  k  d,  will  be  the 
required  elliptical  section.  The  section  upon  the  line 
E  F  may  be  obtained  in  the  same  manner,  all  as  shown 
byepf. 

In  Fig.  62-4,  C  D  F  E  is  a  duplicate  of  the  middle 
piece  of  Fig.  623,  below  which  is  drawn  its  half  plan 
made  up  of  the  elliptical  sections  just  obtained,  all  as 
shown  by  corresponding  letters.  The  piece  thus  be- 


3 , 

^vv___fc__^//^     PLAN          / 


P 

Fig.  624.— The  Middle  Piece  of  Elbow  in  Fig.  623,  with  Plans  of 
its  Bases  Arranged  for  Triangulation. 


comes  an  irregular  flaring  article  or  transition  piece, 
the  envelope  or  pattern  of  which  may  be  obtained  in 
exactly  the  same  manner  as  described  in  Problems  184- 
or  186,  to  which  the  reader  is  referred.  The  eleva- 
tion in  Fig.  623  is  so  drawn  that  C  D  and  E  F  are 
parallel,  and  C  E  is  at  right  angles  to  both.  Should 
the  elevation,  however,  be  so  drawn  that  C  D  is  not 
parallel  with  E  F  the  conditions  will  then  become  the 
same  as  in  Problem  193  succeeding,  which  see;  and 
should  C  E  be  drawn  otherwise  than  vertically,  the 
plan  would  then  resemble  that  shown  in  Prob- 
lem 194. 


PROBLEM   192. 

The  Patterns  for  a  Raking:  Bracket  in  a  Curved  Pediment. 


In  Fig.  625,  let  C  E  F  D  be  the  front  elevation  of 
a  portion  of  a  curved  pediment  whose  center  is  at  K, 
and  of  which  E  K  is  the  center  line.  C  A  B  D  of  the 
same  elevation  represents  the  face  view  of  a  bracket 
having  vertical  sides,  of  which  E  G  F  is  the  normal 


profile.  Since  the  bracket  sides  are  vertical  and  are 
necessarily  at  different  distances  from  the  center  line, 
it  will  be  easily  seen  that  they  are  of  different  lengths 
or  hights ;  that  is,  the  side  C  D,  being  further  from 
E  K  than  the  side  A  B,  is  longer.  The  patterns  for 


Pattern 


357 


the  two  sides  will  therefore  be  different  and  the    face 
piece  will  be  really  an  irregular  flaring  piece. 

It  will  lirst  be  nn-cssary  to  obtain  the  pattern  or 
profiles  of  the  two  sides.  To  facilitate  this  operation 
the  normal  profile  of  the  bracket  E  G  F  has  been  so 


line  E  F,  as  shown.  Thence  carry  lines  around  the 
arch  from  the  center  K,  from  which  the  same  is  struck, 
cutting  the  sides  A  1>  and  C  D.  Conveniently  near 
the  lower  side  of  the  bracket  draw  any  vertical  line,  as 
E1  IV,  as  a  base  line  upon  which  to  construct  the  true 


'-ill       PROFILE 

/12 

13 


RIGHT  SIDE 
ELEVATION 


Fig.  GiTi. — -1  Raking  Bracket  in  a  Curved  Pediment,  Sluttving  the  Patterns  for  It*  Face  and  Sides. 


placed  that  its  vertical  line  or  back  coincides  with  the 
center  line  E  K  of  the  arch.  Divide  the  face  of  this 
profile  into  any  convenient  number  of  parts,  as  shown 
by  the  small  figures,  and  from  these  points  carry  lines 
at  right  angles  to  the  back  of  the  bracket,  cutting  the 


sides  of  the  bracket.  At  any  convenient  position 
upon  this  line,  above  or  below,  as  at  G'  E1  F',  draw  a 
duplicate  of  the  normal  profile,  so  that  its  back  or 
vertical  line  shall  coincide  with  E1  B1,  and  divide  its 
face  line  into  the  same  spaces  as  G  F.  Place  the 


358 


T,,e 


ew 


T-squarc  at  right  angles  to  the  line  E'  B',  and,  bringing 
it  successively  against  tin:  points  in  the  side  0  D, 
draw  lines  cutting  E1  B1,  continuing  the  same  indefi- 
nitely to  the  left,  as  shown.  At  any  convenient  po- 
sition, as  A1  B',  on  the  line  E1  B1  transfer  the  spaces 
from  A  B,  as  shown,  and  from  the  points  thus  ob- 
tained draw  lines  indefinitely  to  the  left  also  at  right 
angles  to  E1  B'.  Place  the  T-square  parallel  to  E'  B', 
and,  bringing  it  successively  to  the  points  in  the  normal 
profile  G'  Fl,  cut  lines  of  corresponding  number  in  the 
two  sets  of  parallel  lines  just  drawn.  Lines  traced 
through  the  points  of  intersection  will  give  the  re- 
quired patterns  of  the  lower  and  upper  sides,  as  shown 
respectively  by  C'  M  D1  and  A1  N  B1.  Some  of  the 
lines  of  projection  in  the  pattern  of  the  upper  side 
have  been  omitted  to  avoid  confusion.  At  the  ex- 
treme left  of  the  engraving  is  shown  a  side  view  of 
the  bracket  as  seen  from  the  right,  which  is  made  up 
of  the  two  sides  just  obtained  and  which  have  been 
placed  in  proper  relation  to  each  other,  all  as  shown 
by  the  dotted  lines  projected  to  the  left  from  the 
points  A,  B,  C  and  D  of  the  front  elevation. 

Having  now  obtained  all  that  is  necessary,  it  re- 
mains to  triangulate  the  face  of  the  bracket  preparatory 
to  developing  the  pattern  of  the  same.  With  this  in 
view  first  connect  all  points  of  like  number  in  the 
upper  and  lower  sides  of  the  front  view  by  solid  lines, 
as  shown.  Also  connect  them  in-the'side  elevation. 
Since  points  of  like  number  in  A  B  and  C  D  have  the 
same  projection  from  the  back  of  the  bracket,  it  will 
be  seen  that  the  solid  lines  just  drawn  connecting 
them  represent  true  distances  across  the  face  of  the 
bracket.  The  four-sided  figures  produced  by  draw- 
ing these  lines  must  now  -be  subdivided  into  triangles 
by  means  of  dotted  lines  drawn  diagonally  through 
each.  Therefore  connect  each  point  upon  the  profile 
of  the  lower  side  of  the  bracket  with  the  point  next 
higher  in  number  upon  the  upper  side,  as  shown  in 
the  side  view.  To  determine  the  true  length  of  these 
lines  it  will  be  necessary  to  construct  a  diagram  of 
triangles,  as  shown  by  S  V  T  in  the  upper  part  of  the 
engraving.  Draw  S  V  and  S  T  at  right  angles  to  each 
other.  Make  S  V  equal  to  the  width  of  the  bracket 
measured  horizontally  across  the  face,  and  upon  S  T, 
measuring  from  S,  set  off  the  lengths  of  the  several 
dotted  lines  in  the  side  view,  as  shown  by  the  small 
figures.  From  each  of  points  in  S  T  draw  lines  to  V. 
Then  these  lines  will  be  the  real  distances  between 
points  of  corresponding  number  on  the  lower  side  of 
the  bracket  and  points  of  the  next  higher  number  upon 


the  upper  side.  The  figures  in  S  T  correspond  with 
the  figures  upon  the.  lower  side  of  the  bracket,  the 
point  V  representing,  in  the  case  of  each  line,  the 
next  higher  number;  thus  -2  V  is  the  distance  from  "2 
to  3'  across  the  face  of  the  bracket,  3  V  the  distance 
3  4',  4  V  the  distance  45',  etc.  The  dotted  lines  in 
the  side  view  representing  the  distances  1  2  and  T  8 
cannot,  of  course,  be  shown  in  that  view,  because 
tliey  lie  in  surfaces  which  appear  in  profile;  but  since 
these  surfaces  are  parallel  with  the  pjane  of  the  back 
of  the  bracket  these  distances  for  use  in  the  pattern 
may  be  taken  directly  from  the  front  view,  as  shown 
by  the  dotted  lines  1  2'  and  7  S'  in  that  view. 

To  lay  out  the  pattern  of  the  face  piece  first  draw 
any  line,  as  C3  A3  or  1  1'  of  the  pattern,  equal  in  length 
to  1  1'  of  the  front  view.  From  C3  of  the  pattern  as 
center,  with  a  radius  equal  to  1  2'  of  the  front  view,  de- 
scribe a  small  arc,  which  intersect  with  another  arc 
drawn  from  A3  as  center,  with  a  radius  equal  to  1'  2'  of 
the  front  view,  thus  establishing  the  point  2'  of  the 
upper  side  of  the  pattern  of  the  face.  From  2'  of  the 
pattern  as  center,  with  a  radius  equal  to  2  2'  of  the 
front  view,  strike  a  small  arc,  which  intersect  with 
another  arc  struck  from  1  of  the  pattern  as  center,  with 
a  radius  equal  to  1  2  of  the  front  view,  thus  establish- 
ing the  position  of  the  point  2  in  the  lower  side  of  the 
pattern.  From  2  of  the  pattern  as  center,  with  a 
radius  equal  to  2  V  of  the  diagram  of  triangles,  strike 
a  small  arc,  which  intersect  with  another  arc  struck 
from  2'  of  the  pattern  of  center,  with  a  radius  equal  to 
2'  3'  of  the  side  view,  thus  establishing  the  point  3'  of 
the  pattern.  From  3'  of  the  pattern  as  center,  with  a 
radius  equal  to  3  3'  of  the  front  elevation,  strike  a 
small  arc,  which  intersect  with  another  arc  .struck  from 
2  of  the  pattern  as  center,  with  a  radius  equal  to  2  3  of 
the  side  view.  So  continue,  using  the  distances  acre  >ss 
the  face  indicated  by  the  dotted  lines  as  found  in  the 
diagram  of  triangles  in  connection  with  the  spaces  in 
the  profile  A"  W  of  the  side  view  to  form  the  upper 
side  A3  B3  of  the  pattern,  and  the  distances  across  the 
face  as  measured  upon  the  solid  lines  of  the  front  view 
in  connection  with  the  spaces  upon  the  profile  C2  1>~  to 
form  the  lower  side  C:i  D3  of  the  pattern, -until  the 
points  13  and  13'  are  reached. 

As  the  lines  C  A  and  D  B  of  the  front  of  the 
bracket  must  be  cut  to  fit  the  curves  of  the  moldings 
above  and  below,  against  which  the  bracket  fits,  the 
corresponding  lines  of  the  pattern  can  be  drawn  with 
radii  respectively  equal  to  K  E  and  K  K,  as  shown  bv 
the  curved  lines  C'  A3  and  D3  B3  of  the  pattern. 


Prwums. 
PROBLEM    193. 

Pattern  for  a  Transition  Piece  to  Join  Two  Round  Pipes  of  Unequal  Diameter  at  an  Angle. 


359 


In  Fig.  626,  DC  Iv  L  shows  u  portion  of  the 
larger  pipe,  of  which  M  P  N  O  is  the  section ;  II  G  B  A 
;i  portion  of  the  smaller  pipe,  of  which  E  J  F  1  is  the 
section;  and  A  B  C  I)  tho  elevation  of  the  transition 
piece  necessary  to  form  a  connection  between  the  two 
pipes  at  the  angle  II  A  L.  The  drawing  also  shows 
that  the  ends  of  the  two  pipes  to  be  joined  are  square, 


Fig.  626.— Elevation  of  a   Transition  Piece  Joining  Two  Round 
Pipes  of  Unequal  Diameter  at  an  Angle. 

or  cut  off  at  right  angles,  so  that  the  lower  base  of 
A  P>  C  D  is  a  perfect  circle  whose  diameter  is  D  C  (or 
M  X )  and  the  upper  base  is  a  circle  whose  diameter. is 
A  B  (or  K  F).  and  also  that  the  side  A  D  is  vertical. 
In  the  choice  of  a  method  of  dividing  the  surface  of 
the  piece  A  BCD  into  triangles,  either  the  elevation 
or  the  plan  can  be  made  use  of  for  that  purpose,  ac- 
cording to  convenience.  In  the  demonstration  here 
given  the  elevation  has  been  used  by  way  of  variety, 


all  as  shown  in  Fig.  627.  Proceed  then  to  divide  the 
plan  of  the  upper  base  into  any  convenient  number  of 
equal  spaces,  as  shown,  and  drop  a  line  from  each  point 
at  right  angles  to  A  B,  cutting  A  B,  and  numbering 
each  point  to  correspond  with  the  number  upon  the 
plan.  In  like  manner  divide  the  plan  of  the  lower 
base  into  the  same  number  of  equal  spaces,  and  erect 
a  perpendicular  line  from  each,  cutting  the  line  D  C, 
and  numbering  the  points  of  intersection  in  the  same 
order,  or  to  correspond  with  the  points  in  the  upper 
base,  all  as  shown.  Connect  the  points  in  D  C  with 
points  of  similar  number  in  A  B  by  solid  lines,  also 


HALF  PLAN  OF 
LOWER  BASE 


Fig.  627. — Triangulation  of  Transition  Piece  Shown  in  Fig.  626. 

connect  points  in  D  C  with  points  of  the  next  higher 
number  in  A  B  by  dotted  lines,  which  will  result  in  a 
triangulation  suitable  for  the  purpose. 

The  next  step  will  be  to  construct  sections  through 
the  piece  upon  all  of  the  lines  upon  the  elevation  (both 
solid  and  dotted),  which  operations  are  shown  in  Figs. 
628  and  629,  and  which  may  be  done  in  the  following 
manner :  Upon  any  horizontal  line,  as  T  S  of  Fig.  628, 
erect  a  perpendicular,  as  T  U.  Upon  T  S  set  off  from 
T  the  several  distances  of  the  points  in  the  lower  base 
from  the  center  line  1  7  of  the  plan,  as  measured  upon 
the  vertical  lines  (Fig.  627),  all  as  indicated  by  the 
small  figures.  Upon  T  U  set  off  the  lengths  of  the 


360 


New  Metal    Worker  Pattern  Bool: 


solid  lines  of  the  elevation,  numbering  eacli  point  thus 
obtained  to  correspond  with  its  line  in  the  elevation. 
From  each  of  the  points  upon  T  U  draw  horizontal 


3    4 

5 


Figr.  628.— Diagram  of  Sections  Fig.  629.— Diagram  of  Sections 
Taken  on  Solid  Lines  of  Fig.  Taken  on  Dotted  Lines  of  Fig. 
627.  627. 

lines  to  the  right,  making  each  in  length  equal  to  the 
distance  of  points  of  corresponding  number  in  the  plan 
of  the  upper  base  from  the  center  line  1  7,  as  measured 
upon  the  lines  at  right  angles  to  line  A  B,  thus  ob- 
taining the  points  2',  3',  4',  etc.  Now  connect  these 
points  with  points  of  corresponding  number  in  the  base 
line  T  S  by  means  of  solid  lines,  as  shown. 

In  constructing  the  sections  upon  the  dotted  lines 
of  the  elevation,  shown  in  Fig.  629,  the  same  course- 
is  to  be  pursued  as  that  employed  in  Fig.  628.  The 
base  line  W  V  is  a  duplicate  of  T  S.  Upon  the  per- 
pendicular line  erected  at  W  set  off  the  lengths  of  the 


Fig.  G31.— Perspective    View  of  Model. 

several  dotted  lines  of  the  elevation,  numbering  each 
point  thus  obtained  to  correspond  with  the  number  at 
the  top  of  its  line  in  the  elevation.  From  each  point 
draw  a  horizontal  line  to  the  right  as  before,  which 
make  equal  in  length  to  the  similar  lines  in  Fig.  628, 
numbering  each  point  as  shown  by  the  small  figures 


2',  3',  4',  etc.  Now  connect  each  of  these  points  with 
the  point  of  next  lower  number  in  the  base  line  V  W 
by  a  dotted  line. 

Having  obtained  all  the  necessary  measurements, 
the  pattern  for  one-half  the  envelope  of  A  B  C  D  may 
be  developed  in  the  following  manner  :  Draw  any  line, 
as  A  D  in  Fig.  630,  which  make  equal  in  length  to  A 
D  of  the  elevation.  With  A  as  a  center,  and  a  radius 
equal  to  1  2  of  the  plan  of  the  upper  base,  Fig.  627, 
strike  a  small  arc,  which  intersect  with  another  struck 
from  D  as  a  center,  with  a  radius  equal  to  the  dotted 
line  1  2'  of  the  diagram,  Fig.  629,  thus  establishing 
the  position  of  the  point  2  in  the  upper  line  of  the  pat- 
tern. From  D  as  a  center,  with  a  radius  equal  to  1  2 
of  the  plan  of  the  lower  base,  Fig.  627,  strike  a  small 
arc,  which  intersect  with  another  struck  from  point  2, 
just  obtained,  with  a  radius  equal  to  the  solid  line  2  2' 


3  4  5 

Fig.  630. — Pattern  of  Transition  Piece  Shou-n  in  Fiy.  G27. 

of  the  diagram,  Fig.  628,  thus  fixing  the  position  of 
point  2  in  the  lower  line  of  the  pattern.  So  continue, 
using  the  lengths  of  the  dotted  lines  in  the  diagram, 
Fig.  629,  in  connection  with  the  lengths  of  the  spaces 
in  the  plan  of  the  upper  base,  to  develop  upper  line  of 
the  pattern,  and  the  lengths  of  the  solid  lines  in  the 
diagram,  Fig.  628,  in  connection  with  the  lengths  of 
the  spaces  in  the  plan  of  the  lower  base,  to  develop  the 
lower  line  of  the  pattern,  using  each  combination  alter- 
nately until  the  pattern  is  complete.  As  each  new 
point  of  the  pattern  is  determined  it  should  be  num- 
bered, and  the  solid  or  dotted  line  used  in  obtaining 
the  same  may  be  drawn  across  the  pattern  if  desired, 
merely  as  a  means  of  noting  progress,  but  these  lines  are 
not  necessary,  as  each  point  is  simply  used  as  a  center 
from  which  to  find  the  next  point  beyond. 

Sometimes,  in  order  to  more  thoroughly  under- 
stand the  method  employed  in  such  an  operation  as  the 
foregoing,  it  is  desirable  to  construct  a  small  model, 


Pattern  Problems. 


361 


which  can  be  made  from  cardboard  or  thin  metal,  the 
details  of  which  are  clearly  shown  in  Fiir.  <>:>1.  The 
pieees  forming  tho  upper  and  lower  bases  of  it  should 
be  duplicates  of  the  half  plans  of  the  upper  and  lower 
bases  shown  in  Fig.  627,  having  the  lines  there  shown 
drawn  upon  them,  and  the  piece  forming  the  back  is  a 
duplicate  of  the  plane  figure  A  B  C  D.  These  three 
parts  may  be  cut  in  one  piece,  after  which  a  right  angle 
bend  on  the  lines  A  B  and  C  I)  will  bring  the  two  bases 
into  correct  relative  position.  Five  quadrilateral  figures 
corresponding  to  those  shown  in  Fig.  G28  may  now  be 


cut  and  fastened  in  position,  according  to  their  num- 
bers, between  the  two  bases  of  the  model.  Threads 
or  wires  can  be  so  plaeed  as  to  correspond  in  position 
with  the  dotted  lines  shown  in  the  elevation,  Fig.  627, 
to  complete  the  model.  The  model  is  only  useful  be- 
fore the  pattern  is  developed  to  assist  in  showing  the 
shapes  and  order  or  rotation  of  the  various  triangles; 
and  one  constructed  to  the  dimensions  of  any  problem 
which  may  occur  to  the  student  at  the  outset  of  his 
study  of  triangulation  will  serve  to  assist  his  imagina- 
tion in  all  subsequent  operations. 


PROBLEM  194. 

The  Pattern  for  a  Flaring  Collar  the  Top  and  Bottom  of  which  are  Round  and  Placed  Obliquely  to 

Each  Other. 


B  3 


Fig.  6SS. -Plan  and  Elevation  of  Flaring  Collar. 


In  Fig.  632  E  F  G  H  shows  the  side  elevation  of 
a  flaring  collar,  the  profile  of  the  small  end  or  top 
being  shown  at  A  B  C  D,  and  that  of  the  bottom  at 
K  L  M  N  of  the  plan.  The-  conditions  embodied  in 
this  problem  are  in  no  respect  different  from  those  of 
the  problem  immediately  preceding.  A  slight  differ- 
ence in  detail  consists  in  the  fact  that  in  the  former 
case  the  short  side  was  at  right  angles  to  the  larger 
end,  while  in  the  present  case  it  is  at  right  angles  to 
the  smaller  end,  but  the  pattern  may  be  obtained  by 
exactly  the  same  method  as  that  employed  in  the 
previous  problem.  However,  as  the  elevation  was 
there  made  use  of  to  determine  the  triangulation,  the 
plan  will  here  be  used  upon  which  to  determine  the 
position  of  the  triangles  of  which  the  pattern  will  sub- 
sequently be  constructed. 

Divide  A  B  C  of  the  profile  into  any  convenient 
number  of  parts,  and,  with  the  T-square  at  right  angles 
to  E  F  of  the  elevation,  carry  lines  from  the  points  in 
ABC,  cutting  E  F,  as  shown.  Extend  the  base  line 
G  II  to  the  left  indefinitely,  and  through  the  center  of 
plan  of  base  draw  0  M,  parallel  with  II  G  of  elevation. 
Drop  lines  from  the  points  in  E  F,  extending  them 
vertically  through  0  M,  With  the  dividers  take  tin- 
distance  across  the  profile  A  B  C  D  on  each  of  the  sev- 
eral lines  drawn  through  it,  and  set  the  same  distance 
off  on  corresponding  lines  drawn  through  0  M.  That 
is,  taking  A  C  as  the  base  of  measurement  in  the  one 
case,  and  O  M  in  the  other,  set  off  on  the  latter,  on 
each  side,  the  same  length  as  the  several  lines  measure 
on  each  side  of  A  C.  Through  the  points  thus  ob- 
tained trace  a  line,  as  shown  by  O  P  Q  R,  thus  obtain- 


TV   Xew  Metal    ll'o/7.r/'    Pattern   Rook. 


ing  the  shape  of  the  upper  outline  as  it  would  appear 
in  the  plan. 

As  both  halves  of  the  plan  when  divided  by  the 
line  0  M  are  exactly  alike,  it  will  only  be  necessary 
to  use  one-half  in  obtaining  the  pattern ;  therefore 
divide  K  N  M  of  plan  into  the  same  number  of  parts 
as  was  the  half  of  profile,  in  the  present  instance  six, 
as  shown  by  the  small  figures.  Number  the  points  in 
K  N  M  to  correspond  with  the  points  in  0  R  Q,  and 
connect  corresponding  points  by  solid  lines,  as  shown 
by  1  1',  2  2',  3  3',  etc.  Also  connect  the  points  in 
O  R  Q  with  those  of  the  next  higher  number  in  K  N 
M,  as  shown  by  the  dotted  lines  1  2',  2  3',  3  4',  etc. 
The  solid  and  dotted  lines  thus  drawn  across  the  plan 
will  represent  the  bases  of  a  number  of  right  angled 
triangles  whose  altitudes  are  equal  to  the  vertical  lines 
between  E  F  and  J  Y  of  the  elevation,  and  whose  hy- 


V  W  X  and  on  W  X  set  off  the  lengths  of  dotted  lines 
iu  plan,  and  from  W  on  W  V  the  lengths  of  lines  in 
E  F  Y  J  of  elevation,  excepting  the  line  E  J,  or  No.  7, 
which  is  not  used.  Connect  the  points  in  W  V  of 
diagram  with  those  of  tire  next  higher  number  in  "W  X, 
as  shown  by  the  dotted  lines  in  diagram  and  by  simi- 
lar lines  in  the  plan.  The  resulting  hypothenuses  will 
give  the  correct  distances  from  points  in  top  of  article 
to-  points  of  next  higher  number  in  the  plan  of 
base. 


.y 

2  • 

3  , 

4 
5 
6 


\\ 
\  \ 


\ 


\ 


43      21U      W 


\ 


2X 


Fig.  633.—  Diagram  of  Triangles 
Based  upon  the  Solid  Lines  of  the 
Plan  in  Fig.  63S. 


Fig.  634.  —  Diagram  of  Triangles 
Based  upon  the  Dotted  Lines  of  the 
Plan  in  Fig.  6S2. 


pothenuses,  when  obtained,  will  give  the  real  distances 
across  the  sides  of  the  finished  article  in  the  direction 
indicated  by  the  lines  across  the  plan. 

To  construct  these  triangles  proceed  as  follows : 
Draw  any  right  angle,  as  S  T  U  in  Fig.  633.  On  T 
U,  measuring  from  T,  set  off  the  lengths  of  solid  lines 
in  plan,  making  T  U  of  diagram  equal  to  Q  M  of  plan, 
T  2  of  diagram  equal  to  2  2'  of  plan,  T  3  of  diagram 
equal  to  3  3'  of  plan,  etc.  From  T  on  T  S  set  off  the 
length  of  lines  in  E  F  Y  J  of  elevation,  making  T  S 
of  diagram  equal  to  F  Y  of  elevation,  T  2  of  diagram 
equal  to  a  2  of  elevation,  T  3  of  diagram  equal  to  b  3 
of  elevation,  etc.  Connect  points  in  T  S  with  those 
of  similar  number  in  T  U,  as  shown  by  the  solid  lines. 
The  hypothenuses  of  the  triangles  thus  obtained  will 
give  the  distance  from  points  in  plan  of  base  to  points 
of  similar  number  of  top  as  if  measured  on  the  fin- 
ished article.  The  diagram  of  triangles  in  Fig.  634  is 
constructed  in  a  similar  manner.  Draw  the  right  angle 


i  m 


Fig.  «.«.—  Pattern  of  Flaring  Collar. 

Having  now  obtained  all  the  necessary  measure- 
ments, the  pattern  may  be  developed  as  follows  :  Draw 
any  line,  as  q  m  in  Fig.  635,  in  length  equal  to  S  U  of 
Fig.  633,  or  F  G  of  elevation.  With  in  of  pattern  as 
center,  and  M  2'  of  plan  as  radius,  describe  a  small 
arc  (2"),  which  intersect  with  one  struck  from  point  (j 
of  pattern  as  center,  and  V  X  of  diagram  of  triangles 
in  Fig.  634  as  radius,  thus  establishing  the  point  2"  of 
pattern.  With  point  2"  of  pattern  as  center,  and  2  2 
of  Fig.  633  as  radius,  describe  a  small  arc  (2),  which 


Pattern  Problems. 


363 


intersect  with  another  struck  from  point  q  of  pattern 
as  center,  and  C  ~2  of  profile  as  radius,  thus  establish- 
ing the  point  2  of  pattern.  With  2  of  pattern  as  cen- 
ter, and  '2  3  of  Fig.  034  as  radius,  describe  a  small  arc 
(3"),  which  intersect  with  another  struck  from  2"  of  pat- 
tern as  center,  and  2'  3'  of  plan  as  radius,  thus  estab- 
lishing point  3"  of  pattern.  With  3"  of  pattern  as 
center,  and  3  3  of  Fig.  033  as  radius,  describe  a 
small  arc  (3),  which  intersect  with  one  struck  from 
point  '2  of  pattern  as  center,  and  the  distance  2  3  of 
prolile  as  radius,  thus  establishing  point  3  of  the  pat- 


tern. Proceed  in  this  manner,  using  the  hypothenuses 
of  the  triangles  in  Figs.  633  and  634  for  the  distances 
across  the  pattern ;  the  distances  between  the  points 
in  the  plan  of  base  for  the  stretchout  of  the  bottom  of 
pattern;  and  the  distances  between  the  points  in  the 
profile  of  top  for  the  stretchout  of  the  top  of  pattern. 
Lines  drawn  through  the  points  of  intersection,  as 
shown  by  q  o  and  m  k,  will,  with  q  m  and  ok,  constitute 
the  pattern  for  half  the  article.  The  other  half  of  the 
pattern  q  o'  k'  m  can  be  obtained  by  any  convenient 
means  of  duplication. 


PROBLEM   195. 

The  Pattern  for  a  Flaring  Flange,  Round  at  the  Bottom,  the  Top  to  Fit  a  Round  Pipe  Passing  through 

an  Inclined  Roof. 


In  Fig.  630,  let  K  L  represent  the  pitch  of  the 
roof,  A  B  C  D  the  elevation  of  the  (hiring  flange,  A  J 
D  the  half  plan  of  the  base,  and  B  F]  C  the  half  plan 
of  round  top  through  which  the  pipe  passes. 


Fig.  U3ti. — Elevation  of  a  Flaring  Flange  to  Fit  Against  an 
Inclined  Rouf. 

It  will  be  seen  by  comparison  that  this  problem 
embodies  exactly  the  same  principles  as  do  the  two 
immediately  preceding,  with  the  slight  difference  in 
detail  that  its  short  side  is  not  at  right  angles  to  either 
upper  or  lower  base.  Also,  in  this  case  the  bottom  of 
the  article  appears  inclined  instead  of  the  top.  It  will 


be  seen  at  a  glance  that  if  the  shape  be  considered  as 
anything  else  than  a  flange  against  an  inclined  roof  the 
drawing  might  be  so  turned  upon  the  paper  as  to  bring 
the  line  K  L  into  a  horizontal  position,  when  it  would 
present  the  same  conditions  as  those  of  Problems  193 
and  194  with  the  slight  difference  in  detail  above  al- 
luded to. 

The  method  of  triangulation  employed  in  this  case 
is  exactly  the  same  as  in  the  problem  immediately  pre- 


\  s 


6   5 


3     2 


Fig.  6S7. — Diagram  of  Triangles  Fig.  638.— Diagram  of  Triangles 
Based  upon  the  Solid  Lines  of  Based  upon  the  Dotted  Lines  of 
the  Plan  in  Fig.  6S6.  the  Plan  in  Fig.  6S6. 


ceding,  and  the  operation  is  so  clearly  indicated  by  the 
lines  and  figures  upon  the  four  drawings  here  given  as 
scarcely  to  need  explanation,  if  the  previous  problem 
has  been  read.  The  plans  of  both  top  and  bottom  are 
divided  into  the  same  number  of  equal  parts,  and  a 
view  of  the  top  as  it  would  appear  when  viewed  at 
right  angles  to  the  base  line  K  L,  and  as  shown  by  F 
G  II,  is  projected  into  the  plan  of  base,  as  indicated 
by  the  lines  drawn  from  B  C  at  right  angles  to  A  D. 


364 


Tlie  New  Metal   Worker  Pattern  Book. 


Points  of  like  number  in  the  two  curves  F  H  G  and  A 
J  D  are  joined  by  solid  lines,  and  the  four-sided  figures 
thus  obtained  are  redivided  diagonally  by  dotted  lines. 
These  solid  and  dotted  lines  become  the  bases  of  the 
several  right  angled  triangles  shown  in  Figs.  637  and 
638,  whose  altitudes  are  equal  to  the  hights  given  be- 
tween the  lines  B  C  and  F  G,  and  whose  hypothenuses 
give  correct  distances  across  the  pattern  between  points 
indicated  by  their  numbers.  The  pattern  is  developed 
in  the  usual  manner  by  assuming  any  straight  line,  as 
C  D  in  Fig.  639,  equal  to  C  D  of  Fig.  636,  as  one  end 
of  the  pattern,  and  then  adding  one  triangle  after  an- 
other in  their  numerical  order;  using  the  stretchout  of 
BEG,  Fig.  636,  to  form  the  upper  line  of  the  pattern, 
the  stretchout  of  A  J  L  to  form  the  lower  side  of  the 
uattern  and  the  various  dotted  and  solid  hypothenuses 


Fig.  639. — Pattern  of  Flaring  Flaitije  Shown  in  Fiy.  636. 

in  Figs.  637  and  638  alternately  to  measure  the  dis- 
tances across  the  pattern. 


PROBLEM    196. 

Pattern  for  an  Irregular  Flaring  Article,  Elliptical  at  the  Base  and  Round  at  the  Top,  the  Top  and 

Bottom  not  Being  Parallel. 


The  conditions  given  in  this  problem  are  essen- 
tially the  same  as  those  of  Problem  193,  but  the  fol- 
lowing solution  differs  from  that  of  the  former  problem 
in  the  method  of  finding  the  distances  from  points 
assumed  in  the  base  to  those  of  the  top,  and  is  intro- 
duced as  showing  varieties  of  method:  In  Fig.  64-0, 
C  G  H  D  represents  the  side  view  of  the  article,  of 
which  G  K  H  is  a  half  profile  of  the  top  and  C  F  D  a 
half  profile  of  the  base.  For  convenience  in  ob- 
taining the  pattern  the  half  profiles  are  so  drawn  that 
their  center  lines  coincide  with  the  upper  and  lower 
lines  of  the  elevation. 

Divide  both  of  the  half  profiles  into  the  same 
number  of  equal  parts — in  the  present  instance  eight. 
From  the  points  obtained  in  the  half  profiles  drop  per- 
pendiculars cutting  G  H  and  C  D.  Connect  the  points 
secured  in  G  H  with  those  in  C  D,  as  a  ?i,  b  m,  etc. 
Also  connect  the  points  in  G  H  with  those  in  C  D,  as 
indicated  by  the  dotted  lines  1  n,  a  m,  etc.  Refer- 
ence to  Problem  193  will  show  that  in  order  to  obtain 
the  correct  lengths  represented  by  the  several  solid 
and  dotted  lines  drawn  across  the  elevation  complete 
sections  upon  those  lines  were  constructed,  as  shown 
in  Figs.  628  and  629.  In  the  present  case  these  dis- 
tances will  be  derived  from  a  series  of  triangles  whose 
bases  are  the  differences  between  the  lengths  of  the 
lines  drawn  across  the  half  profile  of  the  top  and 
those  of  the  bottom. 


Therefore,  to  obtain  the  triangles  giving  the  true  dis- 
tances represented  by  the  solid  lines  proceed  as  follows : 


G, 


Fig.  640. — Elevation  and  Profiles  of  an  Irregular  r  taring  Article, 
Showing  Method  of  Triangitlrttion. 

First  set  off  from  C  D  upon  each  line  in  the  base  C  F  D 
the  length  of  the  corresponding  line  in  the  top ;   thus 


Pattern    Problems. 


365 


make  n  o  equal  to  a  2,  mp  equal  to  I  3,  Z  ?  equal  to  c4, 
etc.  For  the  bases  of  the  triangles  represented  l>v  the 
dotted  lines  set  off  from  C  D  the  length  of  correspond- 


W.. 


r'  q'  i}'  u'  R   u'  t'    s' 


1" 

1 

ll 

a" 

II 

II 

tl 

b" 

II 

II 

II 

c" 

II 

SI 

\ 

II 

\ 

1 

1 

\ 

/ 

c" 

\  \ 
\    \ 

III 
III 

/" 

0' 

:\\ 

\    \  \ 

III 

. 

1     \  ^ 

II  1     ' 

t 

\\    \  \ 

II  1     / 

u  —  WX 

3"2"X"Z"          V  c"7"     5"   4' 


-u 


Representing  Solid  Lines  of  the 
Elevation. 


ing  lines  in  G  K  H,  as  shown  by  the  small  figures  in 
COD.  Thus  make  m  2'  equal  to  a  2,  I  3'  equal  to 
I  :i,  /•  4'  equal  to  c  4,  etc.  The  triangles  represented 
by  solid  lines  in  the  elevation,  and  shown  in  Fig.  641, 
arc  obtained  as  follows:  Draw  the  line  P  Q,  and  from 


641. — Diagram  of  Triangles      Fig.  6J$. — Diagram  of  Triangles 


Representing    Dotted    Lines  of 
the  Elevation. 


from  C  to  0  are  set  off  to  the  left  of  R,  and  the  lengths 
from  0  to  D  on  R  Q.  Thus  make  R  a'  of  diagram 
equal  to  n  a  of  Fig.  640,  R  o  of  diagram  equal  to  o 
17  of  half  profile  of  base,  and  connect  a'  o'.  Make  R 
b'  of  diagram  equal  to  m  iof  Fig.  640,  R  p  of  diagram 
equal  to  p  16  of  the  base,  and  connect  p'  b',  etc. 

The  triangles  represented  by  dotted  lines  in  C  G 
H  D  are  obtained  in  a  similar  manner.  Draw  the  line 
T  U  in  Fig.  642  and  erect  the  perpendicular  V  W. 
From  V,  on  V  W,  set  off  the  lengths  of  dotted  lines 
in  C  G  II  D  of  the  elevation.  Thus  make  V  1"  equal 
to  n  1  of  Fig.  640,  V  a"  equal  to  m.  a,  V  b"  equal  to 

1  6,  etc.     Upon  V  T  or  V  U  set  off  the  lengths  of  the 
lines   in  C  0  D  F  of  Fig.   640,    as    indicated    by    the 
small  figures.     Thus  make  V  X"  equal  to  n  17  of  the 
base  in  Fig.  640,  and  draw  X"  1".     Make  V  2"  equal 
to  2'  16  of  the  base,  and  draw  2"  a",  etc. 

To  obtain  the  pattern  first  draw  the  line  C  G  of 
Fig.  643,  in  length  equal  to  C  G  of  Fig.  640.  From 
C,  with  radius  equal  to  C  17  of  the  half  profile  of 
base,  strike  a  small  arc  (o),  which  intersect  with  an- 
other arc  struck  from  G  as  center,  and  1"  X''  of  Fig. 
642  as  radius,  thus  establishing  point  o  in  the  curve 
of  the  pattern.  From  G  of  pattern  as  center,  and  G 

2  of    the  half   profile  of    top    as  radius,  strike    a  small 
arc,    which   intersect  with  another    arc    struck  from 


\       r 

I            \ 

1           \ 

\ 

\ 

\ 

\ 

\   . 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

Fig.  643.— Pattern  of  Shape  Shown,  in  Fig. 


a  convenient  point  erect  the  perpendicular  R  S,  From 
R  on  R  S  set  off  the  lengths  of  solid  lines  in  C  G  II  D, 
and  from  R  on  P  Q  set  off  the  lengths  of  correspond- 
in<>  lines  in  C  O  D  F,  as  indicated  by  the  small  letters 

o  •* 

ill  COD.      To  avoid  a  confusion  of  lines  the  lengths 


o  of  pattern  as  center,  and  a  o'  of  Fig.  641  as 
radius,  thus  establishing  point  a  in  the  upper  curve  of 
the  pattern.  From  o  of  pattern  as  center,  and  17  16 
in  C  F  us  radius,  strike  a  small  arc  (p),  which  inter- 
sect with  another  arc  struck  from  a  of  pattern  as  cen- 


366 


Tin    .\\-n-  Metal    H'w/'/-   I'utlir 


ter,  and  a"  2"  of  Fig.  64:i  us  radius,  thus  locating 
point  p  of  the  pattern.  From  a  of  the  pattern  as  cen- 
ter, and  23  in  G  K  as  radius,  strike  a  small  arc, 
which  intersect  with  anoth'er  arc  struck  from  />  of  pat- 
tern as  center,  and  p  b'  of  Fig.  64-1  as  radius,  thus 
locating  point  />  of  pattern.  Proceed  in  this  manner. 
using  in  the  order  named  the  spaces  in  the  lower  pro- 
lile,  the  hypotheuuses  of  triangles  in  Fig.  642,  the 


spaces  in  the  upper  prolile.  and  the  hypothenuses  of 
triangles  in  Fig.  <i41.  The  points  thus  obtained,  as 
indicated  by  the  letters  in  Fig.  043,  are  the  points 
through  which  the  pattern  lines  are  to  be  traced.  Then 
C  G  H  D  is  the  required  pattern'  for  one-half  the 
article.  The  other  half  of  pattern,  as  shown  bv  C  G  II' 
D',  can  be  obtained  by  any  convenient  method  of  du- 
plication. 


PROBLEM    197. 

The  Patterns  for  a  Bathtub. 


In  Fig.  644,  let  A  B  C  D  be  the  elevation  and  E 
¥  G  II  half  the  plan  of  a  bathtub.  An  inspection  of 
the  drawing  shows  that  neither  the  segments  forming 


irregular  in  character,  and  the  only  available  method 
by  which  the  various  dimensions  and  curves  constitut- 
ing the  patterns  of  the  same  can  be  ascertained  is  by 


V9 


Fir/.  644. — Plan  and  Elevation  of  Bathtub,  with  Diagrams  nf  Triangles,  Section  of  Top  and  Pattern  of  Sirie  Pier?. 


the  head  nor  those  of  the  foot  of  the  tub  are  concen- 
tric, and  that  their  upper  and  lower  bases  are  not  par- 
allel. Therefore  the  figures  which  they  constitute  are 


dividing  their  surfaces  into  small  triangles,  which  can 
most  easily  be  accomplished  in  the  following  Tnanner: 
Divide  each  of  the  curves  J  I  and  G  II,  forming  the 


Pattern  Problems. 


3t>7 


plan  of  the  head  piece,  into  the  same  number  of  equal 
parts,  numbering  each  the  same,  as  shown,  and  con- 
nect points  of  similar  number  by  solid  lines.  Also 
connect  each  point  in  .J  I,  the  line  of  the  bottom,  with 
tin-  point  of  next  higher  number  in  G  II,  the  line  of  the 
top,  bv  a  dotted  line,  till  as  shown.  The  curves  E  F 
and  L  K,  forming  the  plan  of  the  foot  piece,  are  also  to 
be  divided  into  spaces  and  the  points  connected  by 
solid  and  dotted  lilies  in  the  same  manner  as  those  of 
the  head. 

The  solid  and  dotted  lines  thus  drawn  between  the 
points  in  the  two  curves  of  the  plan  will  form  the 
bases  of  a  series  of  right-angled  triangles,  whose  hypoth- 
enuscs  (after  the  altitudes  are  obtained)  will  give  the 
real  distance  between  the  points  whose  number  they 
bear  upon  the  finished  article.  As,  owing  to  the  slant 
of  the  top  line  A  B  of  the  elevation,  the  triangles  will 


Fig.  645.— Half  Pattern  of 
Head  Plice. 


Fig.  640.— Half  Pattern  of 
Foot  Piece. 


he  of  differing  hights,  the  simplest  way  of  constructing 
them  will  be  as  follows :  Upon  D  C  of  the  elevation 
extended,  as  a  base,  erect  a  perpendicular  line,  M  N 
From  N  on  the  base  line  set  off  the  various  lengths  of 
the  solid  lines  in  the  plan  of  the  head  piece,  as  shown 
toward  (.).  From  each  of  the  points  in  the  curve  G  H 
erect  perpendicular  lines,  cutting  A  B  of  the  eleva- 
tion; and  from  these  points  of  intersection  carry  lines 
horizontally  to  the  right,  cutting  the  line  M  N,  num- 
bering each  point  to  correspond  with,  the  points  in  G  H, 
all  as  shown.  Lines  connecting  points  of  similar  num- 
ber at  M  and  Q  will  be  the  hypothenuses  required,  or 
the  real  distances  between  points  of  similar  number  in 
the  top  and  bottom  of  the  finished  article.  In  a  sim- 
ilar manner  erect  another  perpendicular,  0  P,  and  set 
off  from  P  on  the  base  line  the  lengths  of  the  several 
dotted  lines  in  the  plan  of  the  head  piece,  as  shown 
toward  R.  The  hights  of  the  points  in  the  curve  of 
the  top  can  be  determined  upon  the  line  0  P  by  con- 


tinuing the  lines  drawn  from  A  B  toward  M  till  they 
intersect  O  P.  Each  point  in  the  base  P  R  is  now  to 
he  connected  with  the  point  of  the  next  higher  number 
in  ()  P  by  a  dotted  line.  The  various  hypothenuses 
drawn  between  ()  and  R  will  then  be  the  correct  dis- 
tances between  the  points  connected  by  the  dotted  lines 
of  the  plan.  (The  numbers  in  P  R  correspond  with 
those  in  I  J  of  the  plan  and  not  with  H  G).  The  dia- 
grams from  which  the  dimensions  for  the  foot  piece  are 
obtained  are  shown  at  S  T  U  and  V  W  X  at  the  left 
of  the  elevation  and  are  obtained  in  a  manner  exactly 
similar  to  those  just  obtained  for  the  head  piece, 
all  of  which  is  clearly  shown  by  the  lines  of  the 
drawing. 

From  an  inspection  of  the  drawing  it  will  be  seen 
that  the  line  E  F  G  H  does  not  represent  the  true 
lengths  or  measurements  taken  on  the  top  line  of  the 
tub,  because  A  B,  not  being  horizontal,  is  longer  than 
the  line  E  II,  its  equivalent  in  the  plan,  and  therefore 
a  true  section  on  the  line  A  B  must  be  obtained,  as 
shown  above  the  elevation.  This  may  be  accomplished 
in  the  following  manner :  At  any  convenient  distance 
above  A  B  draw  E'  H1  parallel  to  A  B,  and  from  all 
of  the  points  previously  obtained  on  A  B  carry  lines 
at  right  angles  to  A  B,  cutting  E1  H',  and  extend  them 
beyond  indefinitely,  numbering  each  line  to  correspond 
with  the  point  in  E  F  G  H  from  which  it  is  derived. 
On  the  line  2  set  off  from  E'  H'  a  distance  equal  to 
the  distance  of  point  2  of  the  plan  from  the  line  E  H 
as  measured  on  the  vertical  line;  on  line  3  set  off  as 
before  a  distance  equal  to  the  distance  of  point  3  of 
the  plan  from  E  H,  and  so  continue  until  the  distances 
from  E  H  of  all  the  points  in  E  F  G  H  have  been 
transferred  in  like  manner  to  the  new  section.  Then 
a  line  traced  through  these  points,  as  shown  by  E1  F1 
G1  II1,  will  be  a  section  or  plan  on  the  line  A  B,  from 
which  measurements  can  be  taken  in  developing  the 
upper  edge  of  the  pattern. 

Having  obtained,  by  means  of  the  various  dia- 
grams constructed  in  connection  with  the  elevation,  the 
correct  distances  between  all  the  points  originally  as- 
sumed in  the  plan,  the  pattern  may  now  be  developed 
by  simply  reproducing  all  of  these  distances  or  meas- 
urements in  the  order  in  which  they  occur  upon  the 
plan.  For  the  pattern  of  the  head  piece  assume  any 
line,  as  I  II  of  Fig.  »345,  which  make  equal  in  length  to 
C  B  of  the  elevation,  or  what  is  the  same  thing,  equal 
to  1  1  of  the  diagram  M  N  Q.  From  H  as  a  center, 
with  a  radius  equal  to  1  2  of  the  section  of  top,  strike 
a  small  arc,  which  intersect  with  another  arc  struck 


368 


The  Sew  Mdal   Worker  Put/cm  Book. 


from  I  as  a  center,  with  a  radius  equal  to  1  2  of  the 
diagram  of  dotted  lines  0  P  R,  thus  establishing  the 
point  2'  of  the  pattern.  From  2'  as  center,  with  a 
radius  equal  to  2  2  of  the  diagram  of  solid  lines  M  N  (}. 
strike  a  small  arc,  which  intersect  with  another  struck 
from  point  I  as  a  center,  with  a  radius  equal  to  1  2  of 
the  line  I  J  of  the  plan,  thus  establishing  the  point  2 
of  the  pattern.  Continue  this  operation,  using  in  nu- 
merical order  the  distances  taken  from  the  top  section, 
in  connection  with  the  distances  obtained  from  the 
diagram  of  dotted  lines  O  P  R,  to  form  the  top  line  of 
the  pattern,  and  the  distances  taken  from  the  diagram 
of  solid  lines  M  N  O,  in  connection  with  the  distances 
measured  upon  the  bottom  line  I  J  of  the  plan,  to  form 
the  bottom  line  of  the  pattern,  all  as  indicated  by  the 
solid  and  dotted  lines  drawn  across  a  portion  of  the 
pattern.  Then  I  H  G  J  will  be  one-half  the  pattern  of 
the  head  piece.  The  pattern  for  the  foot  piece  is  de- 
veloped in  exactly  the  same  manner  by  making  E  L 
equal  to  A  D  of  the  elevation,  and  using  the  diagram 
of  dotted  lines  V  W  X  to  measure  upon  the  pattern 
the  distances  indicated  by  the  dotted  lines  upon  the 
plan,  and  the  diagram  of  solid  lines  S  T  U  to  measure 
upon  the  pattern  the  distances  indicated  by  the  solid 


lines  across  the  plan,  the  distances  forming  the  top 
line  of  the  pattern  being  taken  from  E'  F'  of  the  top 
seetiou  while  the  distanced  forming  the  bottom  line  of 
the  pattern  are  taken  from  the  line  L  K  of  the  plan. 

The  pattern  for  the  flat  portion  of  the  side  F  G  J  K 
can  be  obtained  as  follows:  Parallel  to  K  J  of  the 
plan  draw  any  line,  as  K'  J'.  At  right  angles  to  K  .1 
of  the  plan  project  lines  from  points  K.  J,  F  and  G, 
eutting  K1  J',  as  shown,  establishing  the  points  K1  and 
•I',  and  continuing  the  lines  from  points  F  and  G  in- 
definitely. From  K1  of  the  pattern  as  a  eenter,  with  a 
radius  equal  to  8  8  of  the  diagram  S  T  U,  or  of  Fig. 
t>44,  strike  a  small  are,  cutting  the  line  projected  from 
point  F  of  the  plan,  as  .shown  at  F"  of  the  pattern. 
From  J'  of  the  pattern  as  a  center,  with  a  radius  equal 
to  7  7  of  the  diagram  M  N  Q,  or  of  Fig.  644,  strike  a 
small  arc,  cutting  the  line  projected  from  the  point  G 
of  the  plan,  as  shown  at  G2.  Draw  the  lines  K1  F2, 
P  G'J  and  Gs  J1 ;  then  K1  F2  G''  J1  will  be  the  pattern  of 
the  flat  portion  of  the  side. 

The  patterns  of  the  several  parts  can  be  joined 
together,  by  any  convenient  method  of  duplication,  in 
such  a  manner  as  to  produce  as  much  of  the  entire 
pattern  in  one  piece  as  it  is  desired. 


PROBLEM    198. 

The  Pattern  for  a  Flaring:  Flange  to  Fit  a  Round  Pipe  Passing  through  an  Inclined  Roof;  the  Flange 
to  Have  an  Equal  Projection  from  the  Pipe  on  All  Sides. 


In  Fig.  647,  let  a  b  c  d  be  the  elevation  of  the 
pipe,  E  E'  its  plan,  A  B  C  D  the  elevation  of  the 
flange  and  C  D  the  angle  or  pitch  of  the  roof.  Since 
the  projection  of  the  base  of  the  flange  is  required  to 
be  equal  on  all  sides,  as  shown  by  C  1  and  D  1,  the 
flange  will  appear  in  the  plan  as  a  perfect  circle,  F  F'. 
To  avoid  confusion  of  lines  another  elevation  of  the 
flange  G  H  K  J  is  shown  in  Fig.  648,  below  which  is 
drawn  a  half  plan  of  its  base,  MEN,  and  above  which 
is  a  half  plan  of  its  top,  G  L  H,  all  of  which  will  be 
made  use  of  in  dividing  the  surface  of  the  flange  into 
measurable  triangles  for  the  purpose  of  developing  a 
correct  pattern  of  the  same. 

Divide  the  semicircle  G  L  H  into  any  convenient 
number  of  equal  parts — in  the  present  instance  12— 
and  from  the  points  thus  obtained  drop  perpendicular 


lines  to  G  H.  To  obtain  the  shape  of  seetiou  on  roof 
line  J  K  divide  the  half  plan  of  base  MEN  into  the 
same  number  of  equal  parts  as  was  G  L  II,  and  from 
the  points  thus  obtained  carry  lines  at  right  angles  to 
M  N,  cutting  J  K.  From  the  points  in  J  K  draw 
lines  at  right  angles  to  it,  as  shown  by  a  1,  b  2,  c  3, 
etc.  On  these  lines,  measuring  from  J  K,  set  off  the 
length  of  corresponding  lines  in  M  N  B,  thus  making 
lines  a  1,  &  2,  c  3,  etc.,  in  J  K  C  equal  to  lines  «  1,  &2, 
c  3,  etc.,  in  M  N  B.  A  line  traced  through  these 
points,  as  shown  by  J  C  K,  will  give  the  shape  of  sec- 
tion on  roof  and  furnish  the  stretchout  of  base  for 
obtaining  the  pattern. 

In  Fig.  649  is  drawn  a  duplicate  of  the  plan  in 
Fig.  647,  the  spaces  in  its  outer  line  0  1>  P  being 
exact  duplicates  of  the  spaces  in  M  B  N  of  Fig.  648, 


Pattern  1'rublems. 


369 


and  the  spaces  in  its  inner  line  O'  D'  P'  being  dupli- 
cates of  those  in  G  L  II,  all  as  shown  liy  the  small 
liirures.  Draw  solid  lines  connecting  similar  points, 
as  1'  1,  •!'  '1.  .">'  :!.  etc.  In  like  manner  connect  the 

ELEVATION 


PLAN 
Fig.  647. — Plan  and  Elevation  of  Pipe  and  Flange. 

points  in  O'  D'  P'  with  those,  of  the  next  higher  num- 
ber in  0  D  P,  as  0  with  I',  L  with  2',  2  with  3',  etc., 
with  dotted  lines.  These  solid  and  dotted  lines  will 


then  form  the  bases  of  a  series  of  right  angled  trian- 
gles, whose  altitudes  can  be  derived  from  the  eleva- 
tion, and  whose  hypothenuses,  when  obtained,  will  be 
the  correct  measurements  across  the  pattern  between 
points  of  numbers  corresponding  with  the  lines  across 
the  plan. 

To  construct  the  diagrams  of  triangles  represented 
by  solid  and  dotted  lines  in  plan,  extend  G  II  of  Fig. 
•  iJ.s  indefinitely,  as  shown  by  H  W.  From  the  points 
in  J  K  carry  lines  to  the  right  indefinitely,  parallel 
with  G  W,  as  shown  by  the  lines  between  G  W  and 
K  Y.  At  any  convenient  place,  as  R,  and  at  right 
angles  to  GW,  erect  the  lineR  S,  cutting  the  base  line 
K  Y.  From  R  set  off  the  distance  R  T,  equal  to  the 
length  of  any  of  the  solid  lines  in  plan,  Fig.  649,  as 
P'  P,  which  is  the  horizontal  distance  between  the 
pipe  and  lower  edge  of  the  flange.  Draw  T  U  parallel 
with  R  S,  and  also  draw  lines  from  the  points  in  R  S 
to  T.  For  convenience  the  points  in  R  S  can  be  num- 
bered to  correspond  with  the  points  in  J  C  K.  Then 
the  triangle  T  U  S  will  correspond  to  a  section  through 


0  .2 


10 


Fig.  649. — Plan  of  Flange,  Showinij  Trianyulation. 

the  article  on  the  line  P'  P  in  plan,  the  hypothenuse 
S  T  representing  the  distance  between  the  pipe  and 
lower  edge  of  the  flange.  The  diagram  of  triangles  in 
VWYX  is  constructed  in  a  similar  manner ;  draw 
W  Y  at  right  angles  to  G  W,  and  set  off  the  space  W 
V  equal  to  the  length  of  one  of  the  dotted  lines  in 
plan,  Fig.  649,  as  0  1',  and  draw  lines  from  the  points 
in  W  Y  to  V. 

In  developing  the  pattern  the  stretchout  of  top  of 


370 


The  New  Metal    Worker  Pattern,   Book. 


llange  where  it  joins  the  pipe  can  be  obtained  from 
the  semicircle  G  L  II.  The  stretchout  of  lower  edge 
of  flange  where  it  joins  the  roof  can  be  obtained  from 
the  section  on  the  roof  line  J  C  K.  The  distance  be- 
tween points  in  0  D  P  and  0'  D'  P'  of  plan,  Fig.  649, 


length  to  T  S  of  lirst  diagram  of  triangles.  With  the 
dividers  set  to  the  distance  K  1  in  K  C  .1  of  section 
strike  a  small  arc  (!')  from  the  point  <»'  of  pattern. 
With  the  dividers  set  to  the  distance  V  1  of  second  dia- 
gram of  triangles  strike  a  small  arc  from  the  point  0  of 


W 


J 


SECOND 
DIAGRAM 


M  '2 


\ 

Of 

SE 

| 

10 

\ 

)  2 

\ 
e 

Sv 

) 

3 

\ 
e 

^-\^ 

^xX 

4 

6 

B 


My.  648.— Elevation  of  Flanye,  with  Plan,  Sections  and  Diagrams  of  Triangles. 


as  indicated  by  the  solid  lines,  is  given  in  the  diagram 
of  triangles  T  R  S.  The  distance  between  points 
as  indicated  by  dotted  lines  in  plan  is  given  in  the 
diagram  V  "W  Y.  For  the  pattern  then  proceed  as 
follows :  Draw  any  line,  as  IF  K',  Fig.  050,  equal  in 


pattern  as  center,  cutting  the  first  arc  at  1'  of  pattern. 
From  point  1'  of  pattern  as  center,  and  T  1  of  first 
diagram  of  triangles  as  radius,  describe  a  small  arc  (1), 
which  intersect  with  one  struck  from  0  of  pattern  as 
center,  and  0  1  iu  II  L  O  as  radius.  Thus  the  points 


I'ntlrni  Problems. 


371 


'Ml'  and  I  1 '  of  pattern  arc  established.  Proceed  in 
this  manner,  using  in  the  order  described  flic  stretch- 
out obtained  from  the  elliptical  section  K  C  J,  the  hy- 
pothenuse  of  triangle  in  second  diagram  corresponding 
to  the  dotted  line  in  plan,  the  stretchout  from  the  sec- 
tion (r  L  II,  and  the  hypothenuse  of  triangle  in  the 


other  hall  of  the  pattern  can  be  obtained  by  any  means 
of  duplication  most  convenient. 

Should  it  be  required  to  construct  such  a  flange 
to  fit  over  the  ridge  of  a  roof,  it  is  clear  that  that  part 
of  the  flange  shown  in  the  plan,  Fig.  649,  by  O  O'  D'  D 
would  be  a  duplicate  of  the  part  shown  by  D'  P'  P  D, 


.'  12 


J     '*' 


II.  fi.io. — Half  Pattern  of  Flange.  Shown  in  Fig.  (>47. 


first  diagram  corresponding  to  the  solid  line  drawn 
across  the  plan,  until  all  the  measurements  are  used. 
Lines  traced  through  the  points  thus  obtained,  as  shown 
by  H'  G'  and  K'  J',  will  be  the  half  pattern.  The 


and  that,  therefore,  that  portion  of  the  pattern,  Fig. 
650,  shown  by  6  II'  K'  6'  would  be  one-quarter  of  the 
entire  pattern,  which  could  be  duplicated  so  as  to 
make  either  one-half  or  the  whole  pattern  in  one  piece. 


PROBLEM  199. 


Pattern  for  the  Hood  of  a  Portable  Forge. 


In  Fig.  651,  C  A  B  D  represents  the  front  eleva- 
tion of  a  hood  such  as  is  frequently  used  upon  a  por- 
table forge,  K  L  M  N  its  plan  and  E  F  II  J  a  its  side 
view.  The  opening  A  B  at  the  top  of  the  hood  is 
round,  as  shown  by  L  P  of  the  plan,  while  the  base  C  D 
where  it  joins  the  forge  is  nearly  semi-elliptical,  as 
shown. by  K  L  M  of  the  plan.  In  the  side  elevation 
E  «  shows  the  amount  of  flare  and  projection  of  the 


front  of  the  hood,  while  the  opening,  shown  in  the  front 
by  C  S  D,  appears  as  a  simple  straight  line,  a  J.  With 
these  conditions  given,  the  arch  of  the  opening  C  S  D 
of  the  front  elevation  can  be  determined  in  connection 
with  the  plan,  by  projection,  as  shown  by  the  hori- 
zontal dotted  lines,  while  if  the  arch  of  the  front  eleva- 
tion be  assumed  arbitrarily  then  its  line  (a  J)  in  the  side 
view  must  be  obtained  by  projection,  and  will  be  either 


372 


Tim    New 


\\\>rl-rr    Potfrrn    Book. 


straight  or  curved  according  to  the  nature  of  the  curve 
employed  in  the  front  elevation. 

Assuming  the  straight  line  a  J  of  the  side  view  as 
the  true  profile  of  the  arch,  its  curve  in  either  the  front 
elevation  or  the  plan  must  he  determined,  as  a  means 


Fig.  651. — Front  and  Side  Elevations  and  Plan  of  Hwd,  Showing 
System  of  Triangulation. 


of  obtaining  the  pattern.  As  the  flaring  portion  of  the 
hood  very  much  resembles  a  conical  frustum  having 
an  oblique  base,  probably  the  simplest  method  of  ar- 
riving at  its  true  shape  is  to  first  determine  the  plan  of 
this  irregular  frustum  of  which  it  is  a  part.  Therefore 
produce  the  oblique  line  E  a  of  the  side  elevation  until 
it  intersects  the  base  line  H  J  extended  in  the  point 
G.  Next  set  off  from  L  on  the  center  line  of  the  plan 
a  distance  equal  to  H  G  of  the  side  elevation,  thus 
locating  the  point  R.  Through  R,  from  a  center  to  be 
determined  upon  the  center  line,  draw  the  curve  form- 


ing the  front  of  the  plan,  with  such  length  <•!'  radius  as 
will  make  an  easy  junction  with  the  curves  of  the  back 
at  K  and  M.  It  is  not  necessary  that  the  curve  K  R 
M  should  be  a  perfect  circle  throughout;  it  may 
change  as  it  approaches  K  and  M  so  as  to  flow 

smoothly  into  the  as- 
sumed curve  of  the 
back.  It  is  simply 
necessary  that  no 
angle  be  produced  at 
K  and  M, as  such  an 
angle  would  be  con- 
tinued through  the 
surface  of  the  hood 
toward  the  opening 
of  the  top. 

Divide  the  circle 
of  the  top  P  L  into 
any  convenient  num- 
ber  of  equal  spaces, 
as  shown  by  the  small 
figures;  also  divide 
the  outer  curve  of  the 

plan  R  M  L  into  the  same  number  of  spaces.  For  accu- 
racy and  convenience  it  will  be  found  advisable  to  make 
the  spaces  shorter  as  the  curve  increases  from  M  toward 
L  until  the  end' of  the  curve  is  reached  at  the  point  11. 
Connect  points  of  similar  number  in  the  two  curves  by 
solid  lines,  as  shown ;  also  connect  points  in  the  plan 
of  the  top  with  points  of  the  next  higher  number  in 
the  plan  of  the  base  by  dotted  lines.  In  order  to  pro- 
duce the  curve  of  the  opening  correctly  in  the  plan 
and  the  front  elevation  it  will  be  necessary  first  to 
draw  upon  the  side  elevation  lines  corresponding  to 
the  solid  lines  just  drawn  across  the  plan.  To  accom- 
plish this  place  the  "["-square  at  right  angles  to  L  R  of 
the  plan,  and,  bringing  it  successively  against  the  points 
iu  the  plan  of  the  base  R  M  L,  drop  corresponding 
points  on  L  R,  as  shown.  Transfer  the  spaces  thus 
produced  to  the  base  line  H  G  of  the  side  elevation, 
numbering  each  point  to  correspond  with  the  plan. 
By  means  of  the  J-square  placed  as  before,  drop  points 
from  the  plan  of  the  top  to  the  center  line  L  P  (omitted 
in  the  drawing  to  avoid  confusion  of  lines)  and  transfer 
the  same  to  the  line  F  E  of  the  side  elevation,  number- 
ing each  point  as  before.  Now  connect  points  of  cor- 
responding number  in  the  upper  and  lower  lines  of  the 
side  elevation  by  solid  lines,  as  shown;  then  will  those 
lines  be  the  elevations  of  the  solid  lines  drawn  across 
the  plan. 


Pnttirn    1'inlkms. 


373 


It  may  lie  here  remarked  iliat.  as  tin-  pattern  \\-ill 
be  obtained  from  llir  plan,  a  correct  front  elevation  of 
the  opening,  or  arch,  is  not  necessary  t«i  the  work.  Imt 
if  it  is  desired  it  can  be  obtained  in  the  following  man- 
ner: Plaee  the  T-square  parallel  to  L  \i  of  the  plan 
and.  bringing  it  against  the  points  in  the  plan  of  the 


base  between  R  and  M,  drop  corresponding  points  «\i 
the  base  line  C  D  of  the  front  elevation.  Also  in  the 
same  manner  drop  points  from  the  curve  of  the  top  L  P 
in  plan  upon  A  B  of  the  front  elevation,  and  connect 
points  of  corresponding  number  in  the  two  lines  by 
solid  lines,  as  shown.  From  the  points  of  intersection  of 
the  solid  lines  in  the  side  Novation  with  the  line  a  J  (the 
profile  of  the  arch),  a,  b,  c,  etc.,  carry  lines  horizontally 
across,  as  shown,  intersecting  them  with  lines  of  cor- 
responding number  in  the  front  elevation.  A  line 
traced  through  the  points  of  intersection  as  shown  from 
S  to  P,  will  be  the  correct  elevation  of  the  opening  in 
the  front  of  the  hood. 

The  correct  plan' of  the  opening  may  be  obtained  by 
placing  the  T-square  parallel  to  L  K  and  bringing  it 
against  the  various  points  of  intersection  through  which 
the  curve  S  D  was  traced  and  cutting  the  solid  lines 
of  corresponding  number  in  the  plan,  giving  the  points 
a,  6,  c,  etc.  In  case  the  development  of  the  curve  S  D 
has  been  omitted,  measure  the  horizontal  distance  of 
each  of  the  points  «,  &,  c,  etc.,  in  a  J  of  the  side  eleva- 
tion from  the  line  F  II  and  set  off  the  same  on  the 
center  line  of  the  plan  from  L  toward  N.  Thus  the 
horizontal  distance  of  point  a  from  the  line  F  H  is  set 
off  from  L  on  the  center  line  of  the  plan,  thus  locating 
the  point  N  or  a,  the  extreme  point  of  projection  of  the 
hood.  In  the  same  manner  the  projections  of  points 
ft,  c,  etc.,  of  the  side  elevation,  or  in  other  words,  their 
distances  from  F  H  are  set  off  from  L  of  the  plan,  as 


shown  between  X  and  T.  Now  place  the  T"square  a' 
right  angles  to  L  Rand,  bringing  it  against  these  points 
last  obtained,  cut  the  corresponding  solid  lines  of  the 
plan,  thus  locating  the  points  «,  ft,  c,  etc.,  of  the  plan, 
as  before.  A  line  traced  through  these  points  will  be 
the  correct  plan  of  the  curve  of  the  opening. 

Before  the  pattern  can  be  begun  it  will  be  neces- 
sary to  first  obtain  the  correct  distances  represented  bv 
the  solid  and  dotted  lines  across  the  plan.  This  is  ac- 
complished by  means  of  two  diagrams  of  triangles, 
shown  in  Fig.  <>52,  as  follows:  Draw  the  vertical  line 
A  B,  in  length  corresponding  to  the  hight  of  the  hood, 
as  indicated  by  F  II  in  the  side  elevation.  At  right 
angles  to  A  B  draw  B  C,  in  length  corresponding  to  P 
R  or  1  1  of  the  solid  lines  of  the  plan.  From  B  set  off 
also  the  spaces  B  2,  B  3,  B  4,  etc.,  corresponding  in 
length  to  the  lines  2  2,  3  3,  4  4,  etc.,  of  the  plan.  Con- 
nect the  points  in  B  C  with  the  point  A  by  solid  lines. 
Then  will  these  lines  represent  the  true  distances  between 
points  1  and  1,  2  and  2,  etc.,  of  the  plan.  The  second 
diagram  of  triangles  is  constructed  in  a  similar  manner. 
The  vertical  line  E  D  is  drawn,  equal  to  F  II  of  the  side 
elevation.  E  F  is-  set  off  at  right  angles  to  it,  in  length 
equal  to  the  dotted  line  1  2  of  the  plan.  From  E  are 
set  off  the  distances  E  3,  E  4,  etc.,  corresponding  to 
the  lines  2  3,  3  4,  etc.,  of  the  plan.  The  points  thus 
established  in  F  E  are  then  connected  with  D  by  means 
of  dotted  lines.  Then  will  these  lines  represent  the 


Fij,  G-'j-i. — Pattern  of  Hood. 

• 

true  distances  between  points  1  and  2,  2  and  3,  etc.,- 
of  the  plan. 

To  develop  the  pattern,  first  draw  any  vertical  line, 
as    L  Z  of  Fig.   653,   representing  the  center  of   the 


374 


The  N&.O  Metal    Worker  Pattern    Boole. 


bark,  which  make  equal  to  the  liiglit  of  the  hood  I-1  II. 
As  the  base  of  the  hood  is  perfect  ]y  straight  from  L  to 
the  point  11,  set  off  on  a  horizontal  line  from  the  point 
Z,  in  Fig.  653,  a  distance  equal  to  L  11  of  the  plan, 
and  draw  11  L  of  the  pattern.  With  L  as  center,  and 
11  10  of  the  small  circle  in  plan  as  radius,  describe  a 
short  arc.  Then,  from  11  of  the  base  in  pattern  as 
center,  and  11  D  of  the  second  diagram  of  triangles  as 
radius,  describe  a  short  arc  intersecting  the  one  lirst 
drawn,  thus  establishing  the  point  10  of  the  upper  line 
of  the  pattern.  Then  from  this  point  as  center,  with 
A  10  of  the  first  set  of  triangles  as  radius,  describe  a 
short  arc,  and  from  11  of  the  base  of  the  triangular 
portion  of  the  pattern,  with  11  10  of  the  outer  curve 
of  the  plan  as  radius,  describe  another  arc  intersecting 
it,  thus  establishing  the  point  10  in  the  lower  line  of 
the  pattern.  Proceed  in  this  manner,  using  alternately 
the  spaces  in  the  inner  line  of  the  plan,  the  hypothe- 
nuses  of  the  dotted  triangles,  the  hypothenuses  of  the 
triangles  indicated  by  solid  lines,  and  the  spaces  in  the 
outer  line  of  the  plan,  obtaining  the  several  points,  as 
shown.  Then  lines  traced  through  these  points  will 
be  the  pattern  of  the  envelope  of  the  shape  indicated 
by  F  E  Gr  H  of  the  side  elevation,  or  in  other  words, 
of  the  frustum  of  which  the  hood  forms  a  part.  It  now 


remains  to  cut  away  such  a  portum  of  this  pattern  a:: 
represents  the  part  (r  a  J  of  the  side  elevation.  To  ac- 
complish this  it  is  simply  necessary  to  obtain  the  posi- 
tions of  the  points  a,  b,  c,  etc.,  of  the  plan  and  side 
elevation  upon  the  lines  11,22,  3  3,  etc.,  of  the  pat- 
tern. 

With  the  blade  of  the  T-square  set  parallel  to  the 
base  line  G  H  of  the  side  elevation  bring  it  against  the 
points  of  intersection  made  by  the  line  a  J  with  the 
radial  lines,  and  cut  the  vertical  line  F  H,  as  shown  by 
the  short  dashes  drawn  through  it.  Transfer  the  points 
thus  obtained  in  F  II  to  the  vertical  line  A  B  of  the 
first  set  of  triangles.  Then  with  the  blade  of  the  T- 
square  at  right  angles  to  A  B,  and  brought  successively 
against  the  points  in  it,  cut  the  hypothenuses  of  the 
several  triangles  corresponding  in  number  to  the  lines 
from  which  the  points  were  derived  in  the  side  eleva- 
tion, all  as  indicated  by  the  letters  a,  /;,  e,  </,  e  and  /". 
The  distances  of  these  points  from  A  may  now  be 
transferred  to  lines  of  corresponding  number  in  the 
pattern,  measuring  from  the  upper  line,  as  shown  bv 
a,  b,  c,  etc.  Then  a  line  traced  through  these  points, 
as  shown  from  N"  to  M,  will  give  the  shape  of  the  front 
or  arch  of  the  hood,  and  L  P  K  M  Z  will  be  the  half 
pattern  of  the  hood. 


PROBLEM  200. 


The  Patterns  for  the  Hood  of  an  Oil  Tank. 


In  Fig.  654  are  shown  the  elevations  and  plan  of  a 
hood  of  a  style  which  is  usually  hinged  to  the  top  of 
an  oil  tank,  or  can.  The  plan  shows  a  curve  of  some- 
thing more  than  a  semicircle,  H'  Gr  F',  while  the  curve 
F  K  H  of  the  back  view  is  slightly  less  than  a  half 
circle,  the  problem  being  to  determine  the  shape  of  a 
piece  of  metal  to  fill  the  space  between  the  two  curves, 
as  shown  by  A  B  C  of  the  side  view. 

Divide  one-half  of  the  plan  into  any  number  of 
equal  parts,  as  shown  by  the  small  figures  1,  2,  3,  etc. 
From  the  points  established  in  the  plan  carry  lines  up- 
ward until  they  cut  the  base  line  of  the  required  piece, 
as  indicated  by  the  points  between  A  and  B.  From 
the  points  thus  established  carry  lines  parallel  to  A  C 
until  they  cut  the  line  representing  the  back  of  the 
hood,  as  shown  between  C  and  B,  thence  carry  them 


horizontally  until  they  cut  the  profile  of  the  back  of  the 
hood,  as  shown  by  the  points  between  K  and  F.  From 
the  points  in  K  F  drop  lines  vertically  on  to  the  bast- 
line  F  E,  establishing  points  in  it,  as  shown.  Lay  <«|T 
spaces  in  the  line  F'  E'  of  the  plan  corresponding  to 
those  of  F  E  in  the  back,  and  from  the  points  thus 
established  draw  solid  lines  to  those  of  corresponding 
numbers  laid  off  in  the  plan  from  G  to  F'.  These 
lines  represent  the  bases  of  a  series  of  right  angled 
triangles  whose  altitudes  are  shown  by  the  dotted  lines 
of  the  back  view,  and  whose  hypothenuses  will  give 
the  correct  distances  between  points  of  similar  number 
in  the  plan. 

As  the  altitudes  of  these  triangles  are  also  shown 
in  C  B  of  the  side  elevation,  that  view  is  here  made 
use  of  for  the  purpose  of  obtaining  the  required  hy- 


Pattern  Problems. 


375 


pothenuses.     However,    since  tlic  solid    lines   drawn 

across   llic    plan    arc  not   parallel  to  <i    K  .   tin-  distances 
1  B,  2  B,  etc.,  representing  them  in  the  base  line  of  the 


liv  I  lie  solid  lines  drawn  across  the  plan.  To  com- 
plete the  measurements  necessary  for  obtaining  the 
pattern  connect  the  points  in  the  opposite  sides  of  the 
plan  diagonally,  as,  for  example,  0  of 
the  front  and  1  of  the  back,  and  1  of 
the  front  with  2  of  the  back,  as  shown 
by  the  dotted  lines.  These  dotted 
lines  represent  the  bases  of  a  second 
set  of  triangles,  to  be  constructed  in 
the  same  manner  as  the  former  set,  all 
as  shown,  Fig.  655.  Draw  A  B  and  B 
C  at  right  angles  to  each  other  and 
upon  C  B  set  off  the  several  hights 
shown  in  C  B  of  Fig.  654.  Upon  A 
B  lay  off  B  0,  corresponding  in 
length  to  1  0  in  the  plan.  Make  B  1 


h'i(t.i;~>4. — Elevations  and  1'lnn  of  llond  fur  mi  Oil  Tank,  Shnwinit 
System  of  Triangulation. 

side  view  will  not  be  the  correct  bases  of  the  triangles, 
therefore  set  off  on  A  B,  measuring  each  time  from  B, 
the  correct  lengths  of  the  several  solid  lines  of  the  plan, 
as  indicate'1  by  the  points  near  1,  2,  3,  etc.,  on  the  line 

C 


ols         y 

Fiij.  '>•><. — Diagram  of  Trianyles  Based  upon  the  Dotted 
Lines  of  the  Plan. 


A  B,  from  which  points  draw  lines  (shown  dotted)  to 
points  of  similar  number  in  B  C.  Then  the  dotted 
lines  1  1,  ^2,  etc.,  of  the  side  view  will  be  the  correct 
hvpothenuses  of  the  triangles  whose  bases  are  indicated 


of  the  diagram  equal  to  2  1  of  the  plan,  and  in  the 
same  manner  make  B  2  and  B  3  of  the  diagram  equal 
to  3  2  and  4  3  of  the  plan  respectively.  From  the 
points  thus  established  in  the  base  line  of  the  diagram 


Fig.  656. —Pattern  for  the  Top  nf  Hood. 

draw  lines  to  points  of  next  higher  number  in  the  ver- 
tical line.  These  hypothenuses  will  then  represent 
lengths  of  lines  measured  on  the  face  of  the  hood 
corresponding  to  the  diagonal  dotted  lines  in  the  plan. 
To  develop  the  pattern,  first  draw  any  line,  as  0  0 
of  Fig.  656,  equal  in  length  to  A  C  of  side,  Fi«r. 
<i.">4.  From  0,  at  the  right  of  the  pattern,  as  center, 
with  the  distance  between  the  points  0  to  1  in  the  pro- 
file F  K  of  the  back  as  radius,  describe  a  short  arc. 


376 


The  New  Metal    Worker  Pattern    Rook. 


Next  take  in  the  dividers  the  distance  0  1  of  Fig.  655, 
and  from  the  opposite  end  of  the  center  lint-  describe 
a  short  arc,  intersecting  the  one  already  drawn  at  the 
point  1,  thus  establishing  that  point.  From  i  as 
center,  with  dotted  line  1  1  of  the  side  view  as  radius, 
describe  another  short  arc,  which  in  turn  intersect  by 
an  arc  struck  from  0  of  the  left  hand  side  of  the  pat- 
tern with  0  1  of  the  plan  as  radius.  This  will  estab- 
lish the  point  1  of  the  opposite  side  of  the  pattern. 
Continue  in  this  way,  intersecting  the  hypothenuse  of 
the  triangles  whose  bases  are  the  dotted  lines  of  the 
plan  with  the  measurements  taken  from  the  back  view, 


and  the  hypothenuse  of  the  triangles  which  are  shown 
l.iy  the  solid  lines  of  the  plan  with  the  measurements 
taken  from  the  circumference  of  the  plan.  In  this 
manner  all  points  in  the  profile  of  the  pattern  neces- 
sary to  its  delineation  will  be  established.  A  free- 
hand line  drawn  through  these  points  will  give  one 
half  the  required  pattern,  all  as  shown  in  Fig.  fi/iti. 
The  other  half  may  be  obtained  by  any  convenient 
method  of  duplication. 

The  shape  of  patterns  forming  the  back  and  the 
vertical  sides  of  the  hood  are  clearly  shown  in  the 
engraving  and  need  no  further  explanation. 


PROBLEM   201. 

Pattern  for  an  Irregular  Flaring:  Shape  Forming:  a  Transition  from  a  Round  Horizontal  Base  to  a 

Round  Top  Placed  Vertically. 


In  Fig.  657,  let  I  D  E  F  H  represent  the  front 
elevation  of  the  article,  showing  the  circular  opening 
D  E  F  G  forming  its  upper  perim- 
eter or  profile.  The  triangle  A  B 
0  shows  the  shape  of  the  article  as 
it  appears  when  viewed  from  the 
side,  below  which  is  drawn  the  plan, 
showing  its  circular  base,  J  K  L  M. 
The  line  N  P  shows  the  plan  of  the 
opening  D  E  F  G,  which  opening  is 
shown  in  the  side  view  by  that  por- 
tion of  the  line  A  B  from  A  to  Q. 
Opposite  the  front  side  of  the  plan 
N  P  is  drawn  a  duplicate  of  the  pro- 
file D  E  F  G,  as  shown  by  E'  F'  G', 
so  placed  that  its  vertical  center  line 
E'  G'  shall  coincide  with  the  center 
line  of  the  plan,  as  shown.  As  the 
article  consists  of  two  symmetrical 
halves  it  will  only  be  necessary  to 
develop  one-half  the  complete  pat- 
tern. Therefore  divide  one-half  of 
both  profiles  E  F  G  and  E'  F'  G'  into 
the  same  number  of  equal  parts, 
numbering  each  in  the  same  order, 
as  shown  by  the  small  figures ;  also 
divide  the  plan  of  the  base  into  the 
same  number  of  equal  parts  as  the 
profile,  numbering  the  points  to 
correspond  with  the  same.  Drop  lines  from  the  points 
on  the  profile  E'  F'  G'  on  to  the  line  N  P,  at  right  an- 
gles to  the  same,  as  shown,  and  connect  these  points 


with  those  of  similar  number  upon  the  plan  of  base  by- 
solid  lines,  as  shown.      Also  connect  points  upon  the 


4  M 


Fig.  6W. — Elevations  and  Plan  of  an  Irregular  Shape  Forming  a  Transit  inn  from 
a  Round  Horizontal  Base  to  a  Bound  Top  Placed  Verticfi/h/. 


base  with  those  of  the  next  higher  number  upon  the 
line  N  P  by  dotted  lines. 

It  will  be  noticed  that  the  point  J  of    the  plan 


Pattern   Problems. 


377 


represents  at  once  the  point  1  of  tin.-  l>asr  and  the 
points  1  and  7  of  tin:  profile,  shown  by  B,  Q  ami  A 
of  the  side  elevation.  The  linos  drawn  across  the 
plan  represent  the  horizontal  distances  between  the 


3  I  2  B  3  I 

/•'/;/.  U38. — Diagram  of  Triangles. 

points  which  they  connect  and  will  form  the  bases  of 
a  series  of  right  angled  triangles,  whose  altitudes  can 
lie  derived  from  the  elevations,  as  will  be  shown,  and 
whose  hypothenuses,  when  drawn,  will  give  the  true 
distances  between  points  of  corresponding  number 
across  the  finished  article  or  its  pattern.  To  obtain 
the  altitudes  of  the  triangles  carry  lines  from  the 
points  in  the  half  profile  E  F  G  horizontally 
across,  cutting  the  line  A  Q,  as  shown;  then 
the  distances  of  the  points  in  A  Q  from  B 
will  constitute  the  respective  altitudes  of  the 
triangles.  Therefore,  to  construct  a  diagram 
of  all  the  triangles,  draw  any  horizontal  line, 
as  D  C,  Fig.  658,  near  the  center  of  which 
erect  a  perpendicular,  B  A.  Upon  B  A  set 
off  from  B  the  various  distances  from  B  to 
points  in  A  Q  of  Fig.  657,  numbering  the 
same  as  shown  by  the  small  figures.  From 
B  set  off  on  B  C  the  lengths  of  the  various 
solid  lines  drawn  across  the  plan,  Fig.  657, 
and  connect  points  in  B  C  with  those  of 
like  number  in  B  A.  From  B  set  off 
toward  D  the  lengths  of  the  various  dotted 
lines  drawn  across  the  plan  and  connect 
them  by  dotted  lines  with  points  of  the  next 
higher  number  in  the  line  B  A,  all  as  shown ; 
then  these  various  hypothenuses  will  constitute  the 
true  distances  across  the  finished  article  between 
points  of  corresponding  number  indicated  on  the  plan 
and  elevations.  The  distances  between  points  in  the 
base  line  forming  the  larger  or  outer  curve  of  the  pat- 
tern can  be  measured  from  the  base  line  in  plan,  while 


spaces  forming  the  upper  or  shorter  side  of  pattern  can 
be  measured  from   either  of  the  profiles. 

To  develop  the  pattern  it  is  simply  necessary  to 
construct  the  various  triangles  whose  dimensions  have 
been  obtained  in  the  previous  operations, 
beginning  at  either  end  most  convenient 
and  using  the  dimensions  in  the  order  in 
which  they  occur  until  all  have  been  used 
and  the  pattern  is  complete. 

Therefore,  upon  any  straight  line,  as 
A  C  of  Fig.  659,  set  off  a  distance  equal 
to  A  C  of  Fig.  657  or  the  solid  line  7  7 
of  Fig.  658.     From  C  as  a  center,  with 
a  radius  equal  to  7  6  of  the  plan,  Fig.  657, 
describe  a    small  arc  to  the  left,   which 
intersect  with    another   small   arc  struck 
from    A  as  a  center,  and    with  a  radius 
equal  to  the  dotted  line  7  6  of  the  diagram 
of  triangles,  Fig.  658,  thus  establishing  the   position 
of  the  point  6  of  the  pattern.     From  A  of  Fig.  659  as 
center,  with  a  radius  equal  to  7  6  of  the  profile,  Fig. 
657,  describe  a  small  arc,  which  intersect  with  another 
struck  from  point  6  of  pattern  as  center,  with  a  radius 
equal  to  the  solid  line  6  6  of  Fig.  2,  thus  locating  the 
position  of  the  point  6'  of  the  pattern. 


fig.  6Z9.  —  Pattei'ii  of  &it<tjtr  .SV 


in  Fiy.  657. 


So  continue  to  use  the  spaces  of  the  plan,  the 
lengths  of  the  dotted  lines  of  the  diagram  of  triangles, 
the  spaces  in  the  profile  and  the  lengths  of  the  solid 
lines  of  the  diagram  of  triangles  in  the  order  named 
until  all  have  been  used  and  the  pattern  is  complete. 
Lines  traced  through  the  numbered  points  obtained,  as 


378 


T/te  New  Metal    Worker  Pattern   Book. 


shown  from  C  to  B  and  from  A  to  Q,  will  form  the  out- 
lines of  the  pattern  for  half  the  article.     The  other  half, 


A  Q'  IV  C,  can  be  obtained  by  any  means  of  duplica- 
tion most  convenient. 


PROBLEM   202. 


Pattern  for  the  Lining  of  the  Head  of  a  Bathtub. 


In  Fig.  660  are  shown  a  plan  and  side  and  end 
views  of  the  head  of  a  bathtub  or  the  lining  of  a  tub 
the  body  of  which  is  constructed  of  wood.  The  end 
view  shows  the  bottom  corners  of  the  tub  to  be  rounded, 
as  shown  at  C1  G1  and  B1  F1 ;  the  plan  shows  •  the  head 


any  convenient  number  of  equal  spaces,  as  shown  by 
the  small  figures.  In  like  manner  divide  the  quarter 
circle  B'  F'  of  the  end  view  into  the  same  number  of 
equal  spaces,  less  one,  as  also  indicated  by  the  small 
figures.  From  the  points  thus  established  in  B1  F1  carry 


1 

\ 

. 

\ 

. 

\ 

I 

\ 

I 

\ 

\ 

J      END 

VIEW  \ 

\ 

/ 

\ 

m 

\ 

t 

\ 

/ 

\ 

/ 

\      A 

3     3 


Fig.  660. — Elevations  and  Pla,i  of  Lining  for  a  Bathtub,   Showing 
Triangulation  of  the  Head  Piece. 

to  be  semicircular,  while  the  side  view  shows  that 
the  junction  between  the  head  and  the  sides  is 
made  on  the  vertical  line  B"  Ea.  It  will  thus  be  seen 
that  the  conditions  here  given  are  the  same  as  in  the 
previous  problem — viz.,  an  irregular  flaring  piece  form- 
ing a  transition  between  two  quarter  circles  (instead  of 
complete  circles  as  in  the  previous  problem)  lying  in 
planes  at  right  angles  to  each  other. 

Divide  the  quarter  circle  A  F  of  the  top  view  into 


IN 

SECOND  DIAGRAM 


lines  to  the  horizontal  line  B  F  in  the  top  view  and 
mark  the  intersection  by  small  figures,  as  shown.  The 
reason  for  using  one  less  space  in  the  quarter  circle 
B1  F1  than  in  the  large  arc  A  F  is  because  B1  F1  is  not 
the  complete  profile  of  the  end  which  is  to  be  connected 
with  A  F  of  the  top ;  the  line  F1  E  being  required  to 
complete  the  same,  thus  constituting  the  remaining 
space.  Having  established  these  two  sets  of  points  in 
the  plan,  connect  those  of  like  numbers,  as  1  with  1, 
2  with  2,  etc.,  by  solid  lines.  Also  connect  the  points 
in  the  line  of  the  top  with  those  of  the  next  lower 
number  in  the  base,  as  2  with  1,  3  with  2,  etc.,  as  in- 
dicated by  the  dotted  lines.  These  solid  and  dotted 
lines  form  the  bases  of  the  two  sets  of  triangles  shown 
in  the  diagrams  at  the  right,  from  which  the  correct 
measurements  across  the  pattern  are  to  be  obtained. 

To  construct  these  diagrams  extend  A"  E"  of  the 
side  indefinitely  to  the  right,  as  shown.  At  any  con- 
venient points,  as  J  and  M,  drop  the  perpendiculars  J  K 
and  M  N.  From  the  points  established  in  the  quarter 
circle  B1  F1  carry  lines  horizontally  to  the  right,  cut- 


Pattern  Problems. 


379 


ting  the  two  perpendiculars,  as  shown  by  the  small 
figures  above  K  and  N.  From  J,  upon  J  II,  set  off  the 
space  J  1,  equal  to  the  line  1  1  of  the  plan  or  top  view, 
and  from  the  point  1  thus  established  draw  the  hypoth- 
cmise  of  the  triangle,  terminating  in  the  point  1  of  the 
line  J  K.  In  like  manner  set  off  f roin  J,  upon  J  II,  the 
length  2  2  of  the  top  view,  also  3  3,  4  4  and  5  5,  and 
from  the  points  thus  established  draw  lines  to  points 
of  corresponding  designation  in  the  line  J  K,  all  as 
shown.  By  this  means  triangles  have  been  constructed 
the  liypothcniist's  of  which  represent  measurements  on 
the  surface  of  the  finished  article,  taken  on  lines  cor- 
responding to  the  solid  lines  of  the  top  view. 

In  like  manner  construct  the  second  diagram  of 
triangles  shown  at  the  extreme  right,   setting  off  from 


C  B 

f'ii.i.  1:1:1. —J'ntli-rii,  of  Head  FHece  Shown  in  Fig.  660. 

M  distances  equal  to  the  length  of  the  dotted  diagonal 
lines  in  the  top  view,  and  connecting  the  points  thus 
established  with  points  of  next  lower  number  in  the 
line  M  N.  Then  the  hypothenuses  of  this  set  of  tri- 


angles will  give  the  lengths  corresponding  to  measure- 
ments on  the  dotted  lines  of  the  plan  or  top  view. 

Having  now  obtained  all  the  necessary  measure- 
ments, the  pattern  may  be  developed  as  shown  in  Fig. 
661.  The  central  portion  of  the  pattern  will  corre- 
spond to  A'  C'  B'  of  the  end  view,  it  being  simply  a 
flat  triangular  piece  of  metal.  Therefore  draw  any 
horizontal  line,  as  C  B,  equal  in  length  to  C  B  or  C1  B' 
of  Fig.  660.  Take  the  space  1  1  of  the  first  diagrams 
of  the  triangles  as  radius,  and  from  C  and  B,  respect- 
ively, as  centers,  strike  arcs  which  will  intersect  at  A. 
From  A  as  center,  with  1  2  of  the  outer  line  of  the 
plan  as  radius,  describe  a  small  arc.  From  B  as  cen- 
ter, with  1  2  of  the  second  diagrams  of  triangles  as 
radius,  intersect  the  arc  as  shown,  thus  establishing  the 
point  2  in  the  upper  curve  of  the  pattern.  Then  from 
B  as  center,  with  1  2  of  the  arc  B1  F1  of  the  end  view 
as  radius,  describe  another  small  arc,  and  from  2  of  the 
upper  edge  of  the  pattern  as  center,  with  2  2  of  the 
first  diagram  of  triangles  as  radius,  intersect  it  as  shown, 
thus  establishing  point  2  in  the  lower  line  of  the  pat- 
tern. Proceed  in  this  way,  using  alternately  the 
stretchout  of  the  top  of  the  tub,  as  indicated  by  the 
plan  view,  with  the  hypothenuses  of  the  second  dia- 
gram of  triangles  to  establish  the  points  in  the  upper 
curve  of  the  pattern,  and  the  stretchout  of  the  quarter 
circle  shown  in  the  end  view  with  the  hypothenuses 
of  the  first  diagram  of  triangles  to  establish  the  points 
in  the  lower  line  of  the  pattern,  until  the  points  6  and 
5,  or  E  and  F,  are  reached.  Connect  E  and  F  by  a 
straight  line  and  through  the  points  from  A  to  E  and 
from  B  to  F  trace  lines,  as  shown ;  then  A  E  F  B  will 
be  the  pattern  for  one  of  the  corners  of  the  head,  a  du- 
plicate of  which  may  be  reversed  and  transferred  to  the 
other  side  of  the  pattern,  as  shown  by  A  D  G  C,  thus 
completing  the  entire  pattern  of  the  head  in  one  piece. 


PROBLEM  203. 


The  Pattern  for  a  Boss  to  Fit  Around  a  Faucet. 


In  Fig.  662  arc  shown  two  views  of  a  boss  such 
as  is  used  for  fastening  a  faucet  into  the  side  of  a  large 
can  ;  the  curvature  of  the  body  of  the  can  being  rep- 
resented by  the  line  A  D  B.  For  convenience  in 
demonstration,  what  would  be  properly  considered  the 
front  view  of  the  article  is  here  called  the  top  view, 


the  other  view  being  considered  as  the  side.  Let  H  L 
and  K  N  represent  its  desired  length  and  width  of 
base  or  part  to  fit  against  the  body  of  the  can,  and 
PRO  the  circle  of  the  top  to  fit  around  the  neck  of 
the  faucet.  Also  let  D  E  be  its  required  projection 
from  the  can.  Through  E  draw  Y  Z  parallel  to  H  L, 


380 


yy/e   .\(/r   M<:lal    \\~jrlnr   I'nttvrn  JJwk. 


the  long  diameter  of  the  base.  From  1'  ami  O  drop 
lines  at  right  angles  to  H  L,  cutting  Y  Z  in  the  points 
Y  and  Z,  also  from  II  and  L  drop  lines  cutting  A  I)  l>, 
and  connect  the  points  thus  obtained  \vith  Y  and  Z, 
as  shown,  thus  completing  the  side  view. 

Commence  by  dividing  one-quarter  of  the  plan  of 
the  base  K  H  into  any  convenient  number  of  spaces. 
as  shown  by  points  1,  2,  3,  etc.  For  greater  accuracy 
these  spaces  may  be  made  shorter  as  they  approach 
the  ends  of  the  base,  where  the  line  has  more  curve 
than  near  the  middle.  Having  established  the  points 
0,  1,  2,  3,  etc.,  in  K  H,  draw  a  line  from  each  of 
them  to  the  center  of  the  plan  M.  By  this  means  the 
quarter  of  the  circle  representing  the  top  of  the  article, 


measurements,  as  shown  at  the  right  of  C  E,  the 
hypothenuses  of  which  will  represent  the  real  distances 
between  the  required  points. 

Therefore  from  the  points  established  in  II  K 
drop  lines  vertically  cutting  the  section  line  A  .1)  1>, 
as  indicated,  then  carry  lines  from  the  points  on  A  D 
horizontally  till  they  cut  the  Hue  C  K  and  continue 
them  indefinitely  to  the  right.  The  points  at  which 
these  lines  cross  the  center  line  E  C  will  represent  the 
hights  of  the  several  triangles.  On  these  horizontal 
lines,  measuring  fnua  the  center  line  C  E,  which  is 
assumed  as  the  common  perpendicular  for  all  the  tri- 
angles, set  off  the  bases  of  the  several  triangles,  trans- 
ferring the  distances  from  the  plan.  From  the  points 


1         OK 


Y         E          Z 

Fuj.  662. — Top  and  Side  View  of  Boss,  Showing  System  of  Trianyulation. 


and  shown  in  the  diagram  by  P  R,  will  be  divided  in  the 
same  manner  or  proportionately  to  the  plan  of  the 
base,  all  as  shown  by  points  I1,  21,  3',  etc.  It  will  be 
seen  that  these  lines  divide  the  surface  of  the  boss 
into  a  number  of  four-sided  figures,  each  of  which 
must  now  be  redivided  diagonally  so  as  to  form  trian- 
gles. Therefore  connect  0  with  I1,  1  with  2',  etc., 
by  means  of  dotted  lines,  as  shown.  These  solid 
and  dotted  lines  drawn  across  the  top  view  represent 
the  horizontal  distances  between  the  points  given, 
while  the  vertical  distances  between  the  same  can  be 
measured  on  lines  parallel  to  C  E ;  hence  it  will  be 
necessary  to  construct  a  series  of  triangles  from  these 


thus  established  draw  lines  to  E.  which  will  give  the 
hypothenuses  of  the  several  triangles.  For  example, 
on  the  line  drawn  from  the  point  5",  in  A  D  B,  meas- 
uring from  C,  set  off  a  distance  equal  to  5'  5  and 
also  a  distance  equal  to  6'  5  In  the  top  view.  The 
difference  between  these  two  is  so  small  as  to  be  im- 
perceptible in  a  drawing  to  so  small  a  scale  as  this. 
In  like  manner,  on  the  line  drawn  from  4"  set  oil'  a 
distance  equal  to  the  length  of  the  diagonal  lines 
4  5'  and  4  4'  in  the  top  view,  and  in  the  same  manner 
on  the  line  drawn  through  3s  set  off  the  distance  equal 
to  3  41  in  the  top  view  and  also  3  ?>'.  Then,  as  liefore 
remarked,  lines  drawn  from  the  points  thus  established 


Pattern   PrMvnits. 


381 


in  the  horizontal  lines  toward  K  will  be  the  hypoth- 
cnuses  of  tin;  several  triangles  corresponding  to 
sections  represented  by  the  diagonal  lines  in  the  to]> 

view. 

In  view  of  the  fact  that  the  base  of  the  boss  is 
mi-veil  as  shown  bv  A  1)  it  will  be  notieeil  that  the 
measurements  from  K  to  II  in  the  top  view  do  not 
represent  the  real  distances,  because  the  distance  11  M 
is  less  than  the  distance  A  D.  In  ease  extreme  ac- 
curacy is  required  it  will  therefore  be  necessary  to 
develop  an  extended  section  on  the  base  line  A  D, 
\vliich  mav  !"•  done  as  follows:  Kxtend  the  line  M  TT 


required  width  of  the  pattern  on  one  end,  as  shown  by 
Y  6"  in  Fig.  662.  With  these  two  points  established 
proceed  to  obtain  other  points  in  both  lines  of  the 
pattern  by  striking  arcs  with  radii  equal  to  the  spaces 
established  in  the  plan  of  both  base  and  top  of  the 
article  and  to  the  hypothenuses  of  the  triangles  already 
described.  Thus,  from  S  as  center,  with  radius  equal 
to  the  distance  6'  54  of  the  stretchout  of  the  base,  de- 
scribe a  short  arc,  as  shown  at  5  in  the  pattern.  Then 
from  T  as  center,  with  radius  equal  to  E  6'  of  the  tri- 
angles, intersect  it  by  a  second  arc,  as  shown.  From 
T  as  center,  with  radius  equal  to  6'  5'  of  the  plan  of 


TV  U 

Fig.  663.— Pattern  for  Bos*. 


of  the  top  view,  as  shown  at  the  left,  upon  which  place 
a  correct  stretchout  of  A  D ;  that  is,  make  D1  1*  equal 
to  I)  1%  I3  -2*  equal  to  1*  2",  etc.,  and  through  each  of 
the  pofnts  thus  obtained  draw  measuring  lines  at  right 
angles  to  D'  M.  Place  the  T-square  parallel  to  H  M, 
and,  bringing  it  successively  against  the  points  in  the 
line  K  II,  drop  lines  into  the  measuring  lines  of  cor- 
responding number,  as  shown  by  0*,  1',  24,  etc.  Then 
will  the  distances  0*  I4,  1*  24,  etc.,  be  the  correct  dis- 
tances to  be  used  in  developing  the  pattern  instead  of 
the  distances  01,  1  2,  etc. 

The  pattern  may   now  be  developed  as  shown  in 
Fig.  663.      1/iv  oil'  the  line  S  T,  in  length  equal  to  the 


the  top  of  the  article,  describe  a  small  arc,  as  shown,  and 
from  5  of  pattern  as  center,  and  radius  equal  to  E  5'  of 
the  triangles,  intersect  it  by  another  arc,  thus  deter- 
mining the  second  point  in  the  top.  Proceed  in  this 
manner,  adding  one  triangle  after  another  in  the  order  in 
which  they  occur  in  the  top  view,  using  the  spaces  of 
the  plan  of  top  and  of  the  stretchout  of  the  bottom  and 
the  hypotheneuses  of  the  triangles  as  above  described. 
Lines  traced  through  the  points  thus  obtained,  as 
shown  from  S  to  N  and  from  T  to  M,  will  give  the  pat- 
tern of  one-quarter.  This  can  be  duplicated  as  often 
as  is  necessary  to  make  the  entire  pattern  in  one  piece, 
or  to  produce  it  in  halves,  as  shown. 


PROBLEM  204. 


Patterns  for  a  Ship  Ventilator  Having  a  Round  Base  and  an  Elliptical  Mouth. 


In  Fig.  66-1  are  presented  the  front  and  side  ele- 
vations of  a  ship  ventilator  of  a  style  in  common  use. 
A1  B  shows  the  section  or  plan  of  its  lower  piece  A  E 
F  B,  as  well  as  of  the  pipe  to  which  it  is  joined,  while 
K  0  S  P  is  the  shape  of  its  mouth,  or  a  section  upon 


the  line  C  D.  The  curves  E  C  and  F  D  connecting 
the  two  ends  of  the  ventilator  and  forming  the  general 
outlines  of  the  same  may  be  drawn  at  the  discretion  of 
the  designer.  As  the  ventilator  is  constructed  after 
the  manner  of  an  elbow,  it  mav  be  divided  into  as 


382 


T/iu  New  Metut   Worker  Putter n  Bwk. 


many  sections  or  pieces  as  desired.  Therefore  divide 
the  curved  lines  E  C  and  F  D  into  the  same  number 
of  spaces,  and  connect  opposite  points  by  straight 
lines,  as  shown  by  G  II,  K  L  and  M  N.  These  lines 
should  be  so  drawn  as  to  produce  a  general  equality  in 
the  appearance  of  the  different  pieces  without  refer- 
ence to  equality  in  the  spaces  in  either  outline. 

The  next  step  is  to  establish  a  profile  or  section 
upon   each   one  of  these  lines.     These  profiles  can  be 
drawn  arbitrarily,  but  each  should  be  so  proportioned 
that  the  series  will  form  a  gradual  transition  from  the 
circle  A1  B'  to  the  ellipse  R  O  S  P.      All  the  profiles 
will,   therefore,  be  elliptical,   those 
nearer  the  mouth  being  more  elon- 
gated than  those  nearer  the  base  or 
neck.       Since    the   lower   piece    is 
cylindrical  and  is  cut  obliquely   by 
E  F,  the  section  at  E  F  must  neces- 
sarily be  a  true  ellipse  and  can  be 
developed   by  a  method  frequently 
explained  in  connection  with  various 
problems  in  the  first  section  of  this 
chapter,    and  as  also  explained    in 
Geometrical    Problem    68    on  page 
61.    Of  the  remaining  sections,  their 
major  axes  are,  of  course,  equal  to 
the  lengths  of  the  lines  G  H,  K  L 
and  M  N,  and  their  minor  axes  may 
be  determined  by  any  method  most 
convenient,  or  in  the  following  man- 
ner :     Draw  R  U  and  S  V,   repre- 
senting a  front  view    of    the  curved 
lines  passing  through  the  points  n, 
m,  k,  g  and  e  of  the  side  view.    From 
the  points  g,  k  and  m  project  lines 
horizontally  across  to  the  front  view, 
cutting  the  lines  R  U  and  S  V  and  the  center  line  0  T. 
Then  /  I,  d  o  and  b  c  will  be    respectively    one-half 
the  minor  axes  of  the  sections  above  referred  to.    With 
the  major  and  minor  axes  of  the  several  sections  given, 
they  may  be  drawn  by  any  method  producing  a  true 
ellipse,  or  in  case  the  mouth  has  been  drawn  by  means 
of  arcs  of  circles  the  other  sections  may  be  drawn  in 
the  same  manner. 

Each  of  the  several  pieces  of  which  the  ventilator 
is  composed  (except  the  lower  piece)  becomes,  as  will 
be  seen,  a  transition  piece  between  two  elliptical  curves 
not  lying  in  the  same  plane,  and  in  that  respect  is  the 
same  as  the  form  shown  in  Problem  191.  The  pat- 
tern for  each  piece  must,  therefore,  be  obtained  at  a 


separate  operation,  that  for  the  piece  M  N  D  C  only 
being  given.  To  avoid  confusion  of  lines  a  duplicate 
of  it  is  transferred  to  the  opposite  side  of  the  front 
elevation,  as  shown  by  W  Y  Z  X.  Drop  points  from 
Y  and  Z  perpendicular  to  the  center  line  0  T  of  the 
elevation,  thus  locating  the  points  M3  and  N2.  Make 
the  distance  tf  c  equal  to  I  c.  Then  draw  the  ellipse 
M"  I?  N2,  which  will  be  a  front  view  of  the  sec- 
tion M  N  of  the  side  elevation.  On  a  line  parallel 
with  Y  Z  construct  the  section  M'  V  N1,  as  follows: 
Let  M1  N'  be  equal  to  and  opposite  Y  Z.  Let  the 
distance  c  l>'  be  equal  to  the  distance  c  b  of  the  sec- 


w 


— K    •  i  , 


._LYd-_  fffi- 


Fig.  664. — Elevations  and  Sections  of  a  Ship  Ventilator. 


tion.  With  these  points  determined,  draw  through 
them  the  semi-ellipse  M1  V  N'.  Divide  the  sec- 
tions M1  b'  N1  and  0  S  P  into  the  same  number  of 
equal  parts,  as  indicated  by  the  small  figures  in  the 
engraving.  Drop  the  points  1,  2,  3,  4,  etc.,  on  to  and 
perpendicular  to  the  line  Y  Z  ;  thence  carry  them  per- 
pendicular to  the  center  line  O  P  of  the  front  eleva- 
tion, cutting  the  section  M2  b''  N"  in  the  points  l'J,  22, 
3*,  etc.,  thus  dividing  it  into  the  same  number  of 
spaces  as  were  given  to  the  original  section  M1  b1  N1. 
Next  connect  the  points  of  like  numbers  in  the  two 
sections  of  the  front  elevation  by  solid  lines,  thus : 
Connect  2'  with  2",  3'  with  3",  4'  with  42,  etc. ;  also 
connect  the  points  2'  with  1',  3'  with  2',  4'  with  3'J, 


Pattern  Problems. 


etc.,  by  dotted  lines,  all  as  shown  in  the  engraving. 
These  lines  represent  the  bases  of  right  angled  tri- 
angles, whose  altitudes  may  be  measured  on  the  hori- 
zontal lines  cutting  the  lines  W  X  and  V  Z. 

The  next  step,  therefore,  is  to  construct  diagrams 
of  these  triangles,  as  shown  at  A  and  B  of  Fig.  665. 
l>ra\v  any  two  horizontal  lines  as  bases  of  the  triangles, 
and  erect  the  perpendiculars  E  C  and  F  I).  On  both 
K  C  and  F  1)  set:  off  the  various  1  lights  of  the  tri- 
angles, measured  as  above  stated  and  as  indicated  by 
the  points  1,  2,  3,  4,  etc.  Next  set  off  the  length  of 
the  bases  of  the  triangles  as  follows:  In  diagram  A, 
let  C  1  equal  the  distance  1'  T  of  Fig.  664;  make  C  2 
equal  to  2'  2"  and  G  3  equal  to  3'  3",  etc.  Connect 
the  points  in  the  vertical  line  with  the  points  in  the 
horizontal  line  of  the  same  number,  thus  obtaining 


•'  ;;*    D 

Fitj,  665.  —  Diagrams  of  Triangles. 


hypothenuses  of  the  triangles,  or  the  true  distance 
between  the  points  I1  1*,  2'  2",  etc.,  of  the  elevation. 
In  diagram  B,  let  the  distances  D  2,  D  3,  D  4,  etc., 
represent  the  distances  1"  2',  2"  3',  etc.,  of  the  eleva- 
tion. Having  located  these  points,  connect  1  in  the 
vertical  line  with  2  in  the  base;  also  2  in  the  vertical 
line  with  3  in  the  base,  and  proceed  in  this  manner 
for  the  other  points.  This  will  give  the  hypothenuses 
of  the  triangles,  whose  bases  are  1'  2',  2"  31,  etc.,  in 
the  elevation. 

Having  thus  obtained  the  dimensions  of  the  vari- 
ous triangles  composing  the  envelope  of  the  first  section 
of  the  ventilator,  proceed  to  develop  the  pattern  for 
it,  as  shown  in  Fig.  666.  On  any  straight  line,  asC  M, 
set  off  a  distance  equal  to  1  1  in  diagram  A.  From  C 
as  center,  with  radius  equal  to  1'  2'  of  the  elevation, 


Fig.  664,  draw  an  arc,  which  cut  by  another  arc  drawn 
from  M  as  center,  with  radius  equal  to  1  2  of  diagram 
B,  thus  establishing  the  point  2.  From  2  as  center, 
with  radius  equal  to  2  2,  diagram  A,  draw  an  arc, 
which  cut  with  another  arc  drawn  from  1'  as  center, 
with  radius  equal  to  1  2  of  the  elevation,  thus  estab- 
lishing the  point  2'.  Proceed  in  this  manner,  next 
locating  the  point  3,  then  the  point  3' ;  next  the  point 
4,  and  then  4',  etc.  It  will  be  noticed  that,  after 


C  M     - 

Fig.  666.— Pattern  of  First  Section  of  Ship  Ventilator. 

passing  points  6  and  6',  7'  is  obtained  before  7.  This 
is  for  the  sake  of  accuracy,  as  it  will  be  seen  by  in- 
spection of  the  elevation  that  the  distance  7'  6'  is  less, 
and  therefore  more  accurately  measured  in  the  eleva- 
tion, than  the  distance  from  63  to  7'.  Having  thus 
located  the  points  1,  2,  3,  etc.,  I1,  2',  3',  etc.,  trace 
the  lines  C  D  and  N  M,  and  connect  D  with  N,  as  in- 
dicated in  Fig.  666.  Then  D  N  M  C  will  be  the 
pattern  for  one-half  the  section  M  N  D  C  of  the  eleva- 
tion. 

The  pattern  of  the  section  E  A  B  F  will  be  the 
same  as  that  for  the  corresponding  piece  in  an  ordinary 
elbow,  and,  therefore,  need  not  be  specially  explained 
here. 


384:  Tku  Suv  Mdal    Wurkvr  Pattern  Uvok. 

PROBLEM    205. 

Patterns  lor  the  Junction  of  a  Large  Pipe  with  the  Elbows  of  Two  Smaller  Pipes  of  the  Same  Diameter. 


The  elbows  of  the  smaller  pipes  in  the  problem 
here  presented  are  such  as  would,  if  each  were  com- 
pleted independently  of  the  other,  form  six-piece 
elbows.  The  junction  between  the  two  elbows  occurs 
between  the  fifth  pieces,  which  pieces  unite  to  form 
the  transition  from  the  smaller  diameters  of  the  elbows 
to  the  diameter  of  the  larger  pipe,  or  sixth  piece.  A 
pictorial  representation  of  the  finished  work  is  shown 
in  Fig.  667,  in  which,  however,  the  upper  section,  or 
larger  pipe,  is  omitted  to  more  fully  show  the  shape 
and  junction  of  the  transition  pieces.  A  front  view  or 
elevation  of  the  various  parts  is  shown  in  Fig.  608. 
The  side  view  given  in  Fig.  669  shows  more  fully  the 
amount  of  lateral  flare  of  the  transition  piece  necessary 
to  form  a  union  between  the  varying  diameters  of  the 
larger  and  smaller  pipes. 

The  drawing  of  that  portion  of  the  elbows  in  the 
smaller  pipes  from  the  horizontal  parts  up  to  the  line 
a  h  t  in  Fig.  668  is  exactly  the  same  as  that 
employed  in  drawing  a  six-piece  elbow.  The 
piece  A  G  It  «,  occupying  the  place  of  what 
would  otherwise  be  the  fifth  piece  of  the 
elbow,  becomes  in  this  case  an  irregular 
shape,  the  lower  end  or  opening,  «  A,  of  which 
is  nearly  circular  while  its  upper  end,  A  G,  is 
a  perfect  semicircle.  This  piece  unites  with 
its  mate  G  D  t  h  on  the  line  G  //,  thus  form- 
ing the  complete  circle  at  A  D,  a  plan  of 
which  is  shown  immediately  below  the  ele- 
vation. The  relative  proportion  between 
the  diameters  of  the  larger  and  smaller  pipes 
is  such  that  the  junction  between  the  elbows  is  carried 
somewhat  below  the  fifth  pieces,  mitering  the  fourth 
pieces  for  a  short  distance,  as  shown  from  k  to  L.  The 
method  of  cutting  the  lower  parts  of  the  elbow,  how- 
ever, is  the  same  as  that  employed  in  all  elbow  patterns 
where  the  pipe  is  of  a  uniform  diameter  throughout, 
numerous  examples  of  which  are  given  in  Section  1  of 
this  chapter,  to  which  the  reader  is  referred. 

As  the  section  or  profile  of  all  the  parts  forming 
the  elbow  is  a  perfect  circle  when  taken  at  right  angles 
to  the  sides  of  the  pipe,  as  at  Q  F  or  M  N,  it  will  be 
seen  that  a  section  on  the  line  a  It  will  be  somewhat 
elliptical;  it  will  therefore  be  necessary  to  obtain  a 
correct  drawing  of  this  section  from  which  to  obtain 
the  stretchout  of  the  lower  end  of  the  piece  A  G  h  a. 


with  which  it  joins,  and  also  a  drawing  of  it  as  it  will 
appear  in  plan.  Therefore  between  two  parallel  lines 
drawn  from  M  and  N  at  right  angles  to  M  X  construct 
a  profile  or  section,  as  shown  below  at  the  left,  which 
divide  'into  any  convenient  number  of  equal  spaces,  as 
shown  by  the  small  letters  a,  I,  c,  etc.  From  each  of 
these  points  carry  lines  back  to  M  X  at  right  angles  to 
the  same,  and  continue  them  in  either  direction  till 
they  cut  the  miter  line  a  H  of  the  elevation,  as  shown 
by  the  small  letters,  and  the  center  line  n  a  of  the 
section.  From  the  points  in  a  n  of  the  elevation  draw 
lines  at  right  angles  to  the  same  indefinitely,  as  shown 
above  the  elevation,  across  which  at  any  convenient 
point  .draw  a  line,  as  B'  C',  at  right  angles  to  them. 
From  B'  C1  set  off  on  the  lines  last  drawn  distances 
equal  to  the  distances  from  the  circumference  to  the 
diameter  on  corresponding  lines  in  the  section  below, 
all  as  shown  by  a",  i%  c'J,  etc.  A  line  traced  through 


fig.  667.— Perspective  View  of  the  Junction  of  a  Large  Pipe  ivith 
the  Elbows  of  Two  Smaller  Pipes. 


these  points  will  be  the  correct  section  on  the  miter 
line  a  M.  It  will  be  noticed  that  the  section  has  not- 
been  carried  further  than  the  point  It',  the  balance  of 
the  curve  not  being  required  by  reason  of  its  intersec- 
tion with  the  corresponding  piece  in  the  other  elbow. 

Below  the  elevation  and  in  line  with  the  same,  as 
shown  by  the  center  line  G  T,  is  drawn  the  plan  of  the 
larger  pipe  A  B  C  ]).  It  will  be  necessary  to  add  to 
this  the  plan  of  the  curve  on  the  line  «  //  of  the  eleva- 
tion, in  order  that  the  horizontal  distances  between  the 
points  assumed  in  the  two  curves  may  be  accurately 
measured.  Therefore  from  the  points  on  the  miter 
line  a  h  drop  lines  vertically  through  the  plan,  cutting 
the  transverse  center  line  X  Y.  From  X  Y  set  off 
distances  on  these  several  lines  equal  to  the  distances 


Pattern  Probkms. 


385 


of  corresponding  points  from  the  line  a  n  of  the  original 
section,  as  shown  by  «',  //,  etc.,  from  X  to  S.  A  line 
traced  through  these  points  will  give'  the  correct  posi- 
tion of  the  intersection  of  the  smaller  pipe-  as  seen  from 
above.  This  entire,  line  is  shown  in  the  plan,  although 
the  part  from  S  to  X  will  not  be  recpiired,  for  the  reason 
given  above.  An  inspection  <>f  the  plan  will  show  that 
the  side  of  the  plan  from  V  T  to  the  right  would  be  an 
exact  duplicate  of  the  left  side  if  it  were  completed, 
and  that  therefore  the  plan  consists  of  four  symmetrical 
quarters,  one  of  which,  X  R  T,  is  completely  shown 
in  the  plan.  Hence  the  pattern  for  this  quarter 
will  snllice  by  duplication  for  the  entire  transition 
piece. 

Divide  the  quarter  of  the  plan  of  the  larger  pipe 
P  T,  adjacent  to  the  curve  X  S,  into  the  same  number 
of  equal  spaces  as  are  found  in 
the  inner  curve  from  X  to  S, 
as  shown  by  the  small  figures 
1,  2,  3,  etc.  Connect  corre- 
sponding points  in  the  two  lines 
us  shown  by  the  solid  lines  It,' 
8,  ,j'  7,  ./"  li,  etc.  Next  sub- 
divide the  four-sided  figures 
thus  obtained  by  their  shortest 
diagonal,  as  shown  by  the  dot- 
ted lines  ,j  S,/'  7,  etc.  These 
solid  and  dotted  lines  across 
the  plan  represent  the  bases  of 
a  series  of  right  angled  triangles 
whose  altitudes  can  easily  be 
obtained  from  the  elevation, 
and  whose  hypothenuses  when 
obtained  will  give  correct  dis- 
tances across  the  finished  piece 
between  points  connected  on 
the  plan.  These  lines  have 
also  been  drawn  across  the  ele- 
vation from  corresponding  points  in  the  same  for 
illustrative  purposes,  but  such  an  operation  is  not  neces- 
sary to  obtain  the  pattern.  Neither  is  the  side  view 
shown  in  Fig.  669  necessary  to  the  work,  but  is 
here  introduced  merely  to  assist  the  student  in 
forming  a  more  perfect  conception  of  the  operations 
described.  From  the  points  a,  b,  c,  etc.  on  the  miter 
line  a  h  of  the  elevation  carry  lines  horizontally  across, 
cutting  the  vertical  line  G  L,  as  shown  by  the  points 
from  .s-  to  //.  The  distances  of  these  points  from  G  will 
then  represent  the  vertical  distances  of  corresponding 
points  in  X  S  of  the  plan  from  the  plane  of  upper  base 


of  the  transition  piece  shown  by  A  D  of  the  elevation 
and  V  P  T  of  the  plan. 

To  obtain  the  hypothenuses  of  the  various  tri- 
angles above  alluded  to,  or  in  other  words,  the  true 


Fig.  668. — Front  Elevation,  Plan  and  Sections,  Shoiviny  Method  of 
Triangutotion. 

lengths  of  the  lines  dividing  the  surface,  as  shown  in 
the  two  elevations  and  plan,  it  will  be  necessary  to 
construct  a  series  of  diagrams,  as  shown  in  Fig.  670. 
Therefore  draw  any  two  lines,  as  A  S  and  h  h',  at  right 


386 


The  Xt-iv  Metal    Worker  Pattern  Bwk. 


angles  to  each  other;  make  h  8  equal  to  h'  8  of  the 
])lan,  Fig.  668,  and  h  h'  equal  to  h  8  of  the  elevation, 
and  draw  h'  8.  Next  draw  any  two  lines,  as  rj  8  and  <j  7', 
at  right  angles  to  each  other,  making  g  8  equal  to  the 


Fiy.  669.— Side  Elevation. 

dotted  line  g'  8  of  the  plan  and  g  7  equal  to  the  solid 
line  g'  1  of  the  plan.  Make  g  g'  equal  to  the  distance 
of  point  g  from  the  line  A  D  as  measured  by  its  cor- 
responding point  on  the  line  L  G.  Draw  g  8  and  g  7. 
So  continue  till  all  the  triangles  have  been  constructed. 
Then  the  solid  hypothenuses  will  represent  the  true 
distances  across  the  pattern  indicated  by  the  solid  lines 
of  the  plan  and  elevations,  and  the  dotted  hypothenuses 
the  true  distances  on  corresponding  dotted  lines  in 


make  equal  to  the  line  1  a'  of  the  diagram  of  triangles, 
Fig.  670.  From  1  as  a  center,  with  a  radius  equal  to 
1  2  of  the  plan,  describe  a  small  arc,  which  intersect 
with  another  small  arc  drawn  from  a  as  center,  with  a 
radius  equal  to  a'  2  of  the  diagram  of  triangles,  thus 
locating  the  point  2  of  the  pattern.  From  point  '2  as 
a  center,  with  a  radius  equal  to  b'  2  of  the  diagram  of 
triangles,  describe  a  small  arc,  which  intersect  with  an- 


Fiij.  071. — Pattern  for  One-Quarter  of  Uunitcctiag  Piece. 

other  small  arc  struck  from  a  of  the  pattern  as  center, 
with  a  radius  equal  to  a'  I?  of  the  section  on  line  "  // 
of  elevation  shown  above,  thus  establishing  the  posi- 
tion of  the  point  b  of  the  pattern.  Proceed  in  this 
manner,  using  the  spaces  in  P  T  in  the  plan  of  the 
larger  pipe  to  form  the  upper  edge  of  the  pattern  and 
the  spaces  from  the  section  B1  C1  to  form  the  lower 
edge  of  the  pattern,  measuring  the  distances  between 
the  same  by  the  alternate  use  of  the  solid  and  dotted 
hypothenuses  of  corresponding  number  and  letter  taken 
from  the  diagram  of  triangles  in  Fig.  670.  A  line 
traced  through  the  two  series  of  points  and  a  straight- 
line  from  8  to  h  will  complete  the  pattern  for  one- 


g  '  -?- 


«T  4    5  e       5  e 


Fit/.  670.—Dianram  of  Triawjles. 


those  views.     In  describing  the  pattern,  work  can  be 
begun  at  either  end  of  the  pattern  most  convenient. 
Praw  any  straight  line?  as  1  a  of  Fig.  671,  which 


quarter  of  the  transition  piece  required.  The  remain- 
ing three-quarters  can  be  obtained  by  any  means  of 
duplication  most  convenient. 


387 


PROBLEM   206. 

The  Patterns  for  a  Right  Angle,  Two-Piece  Elbow,  One  End  of  Which  is  Round  and  the  Other  Elliptical. 


In  Fig.  072,  lot  A  G  C  B  H  D  represent  the  eleva- 
tion of  elbow.  A  F  D  the  half  profile-  of  elliptical 
end  and  C  K  B  the  half  profile  oC  round  end.  The 
first  step  will  be  to  establish  a  section  on  the  miter 
line  G  II.  Since  the  width  at  A  1).  one-half  of  which 
is  shown  by  K.  F,  is  greater  than  .1  K.  one-half  the 
width  at  C  B,  it  is  proper  that  the  width  at  L  should 
be  a  medium  between  the  two.  Therefore  from  K, 
on  K  F,  set  off  the  distance  J  K,  as  indicated  by  K  m. 
Bisect  F  771  in  «,  and  take  K  n  as  the  width  at  L.  The 
section  at  G  H  will  then  be  an  ellipse,  of  which  G  ][ 
is  the  major  axis  and  K  n  one-half  of  the  minor  axis. 


fig.  672.— Elevation   of  Elboir,  with  Half  Profiles  of  the  Two  Ends. 

In  Fig.  673,  A  G  II  D  is  a  duplicate  of  the  part 
bearing  the  same  letters  in  Fig.  672.  Against  A  D 
is  placed  a  half  profile,  A  F  I),  of  the  elliptical  end. 
while  against  G  H  is  placed  one-half  of  the  elliptical 
section,  constructed  as  above  described  and  as  shown 
by  G  L  H.  Divide  G  L  H  into  any  convenient  num- 
ber of  equal  parts,  and  from  the  points  thus  obtained 
drop  perpendiculars  cutting  G  II,  as  shown.  Also 
divide  A  F  D  into  the  same  number  of  parts,  and  from 
the  points  thus  obtained  drop  perpendiculars  cutting 
A  D.  Connect  the  points  in  A  D  with  those  in  G  H, 
as  indicated  by  the  solid  and  dotted  lines. 

The  next  step  is  to  construct  sections  on  each  of 


the  -solid  and  dotted  lines  drawn  across  the  elevation 
by  means  of  which  to  obtain  the  true  distances  between 
the  points  in  A  D  and  those  in  G  II  as  though  meas- 
ured upon  the  finished  article.  In  Fig.  674  is  shown 
a  diagram  containing  sections  upon  the  solid  lines, 
which  is  constructed  in  the  following  manner :  Draw 
aii\  two  lilies,  as  M  N  and  M  P,  at  right  angles  to  each 
other.  Upon  M  N  set  off  the  bights  of  the  several 
points  in  the  profile  A  F  D;  thus  make  M  13,  M  12, 
Mil,  etc.,  respectively  equal  to  k  13,  j  12,  A  11,  etc., 
of  Fig.  673.  Upon  M  P  set  off  from  M  the  lengths 


/•'if/.  673.— Elevation  of  Lower  Piece  of  Elbow,   Showing  Method 
of  Triftnyulation. 

of  the  several  solid  lines  of  the  elevation ;  thus  make 
M  a,  M  b,  etc.,  respectively  equal  to/  a,  g  b,  etc.,  of 
Fig.  673,  and  at  the  points  a,  b,  c,  etc.,  thus  obtained, 
erect  perpendiculars,  each  equal  in  hight  to  the  bight 
of  the  corresponding  point  in  the  profile  G  L  II  of 
Fig.  673  from  the  line  G  II.  Thus  make  a  2,  b  3, 
etc.,  of  the  diagram  respectively  equal  to  a  2,  b  3, 
etc.,  of  Fig.  673,  and  from  the  points  2,  3,  4,  etc., 
thus  obtained,  draw  solid  lines  to  points  9,  10,  11, 
etc.,  in  the  line  M  N,  all  as  shown.  Then  the  dis- 
tances 9  2,  10  3.  11  4,  etc.,  will  be  the  true  lengths 
represented  by  corresponding  solid  lines  drawn  across 
the  elevation. 


38$ 


flte  New  Metal    Worker  Pattern  Book 


The  true  distances  represented  by  the  dotted  lines 
drawn  across  the  elevation  are  obtained  in  the  same 
manner  by  means  of  the  diagram  shown  in  Fig.  ti7r>. 
E  S  is  drawn  at  right  angles  to  E  T  and  upon  it  are 
set  off  the  bights  of  the  points  in  A  F  D  the  same  as 
in  M  N  of  Fig.  674.  Upon  E  T  set  off  from  B  the 
lengths  of  the  several  dotted  lines  drawn  across  the 


11 

. 

.  ; 

: 

10,12 

O 

I 

8,13 

.  " 

- 

M 

a       b         c        d       e 
Fig.  674.— Diagram  of  Sections  Upon  Solid  Lines  of  Fig.  673. 

elevation,  as  shown  by  corresponding  letters,  and  from 
the  points  thus  obtained  erect  perpendiculars  also  as 
in  Fig.  674.  Finally  connect  by  dotted  lines  such 
points  as  correspond  with  those  connected  by  dotted 
lines  in  the  elevation.  Thus  from  9  in  E  S  draw  a 
line  to  point  1  in  the  base  line,  corresponding  to  the 
line  /  1  of  the  elevation,  Fig.  673.  Lines  from  10  to 
2  and  from  11  to  3  of  the  diagram  will  correspond 
respectively  to  g  a  and  h  b  of  Fig.  673. 

To  develop  the  pattern  from  the  dimensions  now 
obtained  proceed  as  follows :    At  any  convenient  place 


from  TT  of  pattern  as  center,  describe  an  arc.  whidi 
cut  with  another  arc  struck  from  point  9  of  pattern  as 
center,  and  9  ~2  of  Fig.  IJ74  as  radius,  thus  establish- 
ing point  •>  of  pattern.  With  point  2  of  pattern  as 
center,  and  2  10  of  Fig.  675  as  radius,  describe  an 
arc,  which  intersect  with  another  arc  struck  from  point 
9  of  pattern  as  center,  and  9  10  of  profile  as  radius, 


11 

^-^^                              ' 

10,12 

C: 

3 

.--• 

5 

^-'*"1*: 

•> 

9,13 

X 

^ 

."*" 

N 

* 

•""" 

_N         fff.  ' 

R  lab         c       d      e 

Fig.  r,75.— Diagram  of- Sections  Upon  Dotted  Lines  of  Fig.  673. 

thus  establishing  point  10  of  pattern.  With  point  10 
of  pattern  as  center,  and  10  3  of  Fig.  674  as  radius, 
describe  a  small  arc,  which  intersect  with  one  struck 
from  point  2  of  pattern  as  center,  and  2  3  of  G  L  II  as 
radius,  thus  establishing  point  3  of  pattern.  Continue 
this  process,  locating  in  turn  the  remaining  points  in 
pattern,  as  shown.  Lines  drawn  through  the  points 
thus  obtained,  as  indicated  by  G  II  and  D  A,  will  be 
one-half  of  the  required  pattern.  The  other  half  of 
the  pattern  can  be  obtained  in  a  similar  manner, 
or  by  tracing  and  transferring.  The  pattern  for  the 


Fig.  676.— Pattern  for  Lower  Piece  of  Elboiv. 


draw  the  straight  line  H  D  in  Fig.  676,  in  length  equal 
to  H  D  of  Fig.  673.  From  II  of  pattern  as  center, 
with  radius  1  9  of  Fig.  675,  describe  an  arc,  which 
intersect  by  a  second  arc  struck  from  D  as  center,  with 
radius  D  9  of  profile,  thus  establishing  the  point  9  of 
pattern.  Then  with  radius  II  2  of  G  L  H,  Fig.  673, 


other  part  of  elbow,  as  shown  in  Fig.  672  by  G  C  B  H. 
can  be  obtained  by  the  same  method.  The  shape 
G  L  H  of  Fig.  673  is  to  be  drawn  to  the  left  of  the 
miter  line  G  H,  and  the  operation  continued,  usiiiL: 
the  same  process,  as  shown  in  Figs.  674,  675  and 
676. 


Pattern    Problems. 


389 


PROBLEM    207. 

The  Pattern  for  a  Y  Consisting  of  Two  Tapering  Pipes  Joining  a  Larger  Pipe  at  an  Angle. 


In  Fig.  677,  BCD  K  represents  ilie  elevation  of  a 
portion  of  the  larger  pipe  and  C'  K  D'  L  its  prolile. 
This  pipe  is  cut.  oil:  square  at  its  lower  end,  witliwliieh 
the  liranches  of  the  Y  are  to  be  joined.  A  15  O  II  J 
and  G  II  0  E  F  are  the  elevations  of  the  two  similar 
tranches  joining  each  other  from  II  to  0,  and  the  larger 


l-'iii.  >!77. — Elevation  find  Profiles  nf  Y  with  Tapering  Branches. 


pipe  on  the  line  B  E.  A'  N  J'  M  is  the  profile  of  one 
of  the  tapering  branches  at  its  smaller  end. 

Since  the  article  consists  of  two  symmetrical  halves 
when  divided  from  end  to  end  on  the  lines  A'  J'  or  C' 
D'  of  the  profiles,  and  since  the  two  branches  are  alike, 
the  pattern  for  one-half  of  one  of  the  branches,  as  A  B 
()  II  J,  is  all  that  is  necessary. 

The  dividing  surface  A  B  0  II  J,  lying  as  it  were 
at  the  back  of  the  half  of  the  branch  shown  in  eleva- 


tion by  the  same  letters,  will  then  form  a  plane  or  base 
from  which  the  hights  or  projection  of  all  points  in  the 
surface  of  the  branch  piece  can  be  measured. 

As  the  branch  piece  A  B  0  H  J  is  an  irregular 
tapering  form,  its  surface  must  be  divided  into  a  series 
of  measurable  triangles  before  its  pattern  can  be  ob- 
tained. Therefore  divide  the  half  profile  C'  L  D'  into 
any  convenient  number  of  equal  parts — in  the  present 
instance  six,  as  shown  by  the  small  letters  fghj  k — and 
from  these  points  drop  lines  parallel  with  C  B,  cutting 
the  line  B  K,  as  shown.  In  a  similar  manner  divide 
the  half  profile  A'  N"  J'  into  the  same  number  of  equal 


Fig.    678. — Elevation   of  One  Branch  of  Y,   Shouting  Method  of 
Triangvlation. 

parts  as  was  C'  L  D',  as  shown  by  the  small  letters  a 
i  bed  e.      From  the  points'  thus  obtained  carry  lines  par- 
allel with   J'  J,  cutting  A  J.      Connect  the   points  in 
A  J  with  those  in  B  E,  as  shown. 

To  avoid  a  confusion  of  lines  the  subsequent  oper- 
ations are  shown  in  Fig.  678,  in  which  A  B  0  II  .1  is  a 
duplicate  of  the  piece  bearing  the  same  letters  in  Fig. 
677.  The  profiles  B  L  K  and  A  X  .1  are  also  dupli- 
cates of  those  shown  in  Fig.  677  and  arc  for  conven- 
ience here  placed  adjacent  to  the  lines  which  they  rep- 
resent. B  L  of  the  upper  profile  then  represents  a 
section  on  the  line  B  0.  and  A  N  J  that  upon  the  line 
A  J,  but  the  section  on  the  line  0  H,  the  miter  be- 
tween the  two  branches,  is  as  yet  unknown.  To  obtain 
this  it  will  be  necessary  to  first  obtain  sections  upon 


390 


New  Metal    Worker  Pattern  Book. 


the  various  lines  drawn  across  the  elevation  from  B  V, 

to  A  J  in  Fig.  678,  or   in  other  words,  diagrams  upon 

which  the  true  lengths  of  those  lines  can  be  measured. 

In  the   diagram   of    sections   shown  in  Fig.   679 


fig.  680. 
Diagrams  of  Sections  Upon  Solid  Lines  of  the  Elevation. 

S  T  represents  the  dividing  surface  or  base  plane  al- 
luded to  above  and  is  made  equal  in  length  to  2  2'  of 
Fig.  678.  At  either  extremity  of  this  line  erect  the 
perpendiculars  S b  and  T/,  as  shown.  Make  T/ equal 
in  hight  to  2' /of  profile  B  L  E,  and  upon  S  b  set  off 
from  S  the  hight  S  a,  equal  to  2  a  of  the  profile  A  N  J, 
and  draw  the  line  a  /.  On  S  T,  measuring  from  T, 
set  off  the  distance  2'  2",  and  erect  the  perpendicular 
2"/",  cutting  af  at  f".  Then  will  a  f  represent  the 
true  distance  between  the  points  2  and  2'  in  Fig.  678, 
and  af"  will  represent  the  true  distance  from  2  to  2",- 


<r 


T? 3* 

x 


(?.  681. 


u 


7' 6'    5' 
V 


Fijr.  683. 
Diagrams  of  Sections  Upon  Dotted  Lines  of  the  Elevation. 

while  2"  /"  will  be  the  hight  of  the  point  2".  In  a 
similar  manner  set  off  from  S,  on  S  T,  a  distance  equal 
to  3  3'  of  Fig.  678  and  erect  the  perpendicular  3'  g, 
equal  in  length  to  3'  g  of  profile  B  L  E.  Make  S  b  equal 
to  3  b  of  profile  A  N"  J  and  draw  b  g.  From  S  set  off 


on  R  T  a  distance  equal  to  3  3"  of  Fig.  67s  and  erect 
the  perpendicular  :!"  y",  cutting  b  rj  at g".  Then  will 
l>  '/  be  equal  to  the  true  distance  between  3  and  3'  of 
Fig.  678,  b  g"  will  be  the  true  distance  from  3  to  3"  of 
Fig.  678  and  3"  g"  will  be  the  hight  of  the  point  3".  To 
construct  the  section  on  the  line  0  II,  at  points  3"  and 
2 "draw  3" y'  and  2"/'  at  right  angles  to  0  II,  making 
them  respectively  equal  ?,"  y"  and  2"/"  of  Fig.  679. 
As  the  profile  B  L  E  is  a  semicircle  the  hight  of  point 
4' — that  is,  4'  h — is  equal  to  0  E  ;  therefore  through  the 
points  E,  «/',/'  and  II  draw  the  curve  sliown,  which 
will  be  the  true  section  on  line  0  II,  from  which  the 
stretchout  can  be  taken  for  that  portion  of  the 
pattern. 

The  sections  on  the  remaining  lines  (4  4',  5  5'  and 
6  6')  of  the  elevation  are  shown  in  Fig.  680  and  are 
constructed  in  exactly  the  same  manner  as  those  shown 
in  Fig.  679,  giving  c  Ji,  dj  and  e  k  as  the  true  lengths 


O1 


iii.  (183.  —  Pattern  of  Tapering  Branch. 


of  those  lines.  Before  the  pattern  can  be  developed 
the  four-sided  figures  into  which  the  surface  of  the 
branch  pipe  has  been  divided  by  the  solid  lines  must 
be  subdivided  into  triangular  spaces,  as  sliown  by  the 
dotted  lines  in  the  elevation.  Sections  upon  these  lines 
must  also  be  constructed,  in  order  that  their  true 
lengths  can  be  obtained.  These  are  shown  in  two 
groups  in  Figs.  681  and  682  and  are  constructed  in  a 
manner  exactly  similar  to  that  described  in  connection 
with  Fig.  679.  They  may  be  easily  identified  by  cor- 
respondence between  the  figures  on  the  base  lines  V  V 
and  W  X  and  those  of  the  elevation. 

To  describe  the  pattern  proceed  as  follows  :  Draw 
any  line,  as  J  H  in  Fig.  683,  in  length  equal  to  J  H  of 
Fig.  678.  With  J  of  pattern  as  center,  and  J'  a  of 
smaller  profile  as  radius,  describe  a  small  arc  (a),  which 


391 


Cut  with  one  struck  from  IF  of  pattern  as  center,  and 
1"  a  of  Ki^.  lixl  as  ,-ailius.  thus  establishing  the  point 
a  of  pattern.  With  u  of  pattern  as  center,  and  n  /"'of 
Fig.  679  as  radius,  describe  another  small  are  (/'), 
which  intersect  with  one  struck  from  II  of  pattern  as 
center,  and  II /'  of  profile  II  K  as  radius,  thus  estab- 
lishing the  point  /'of  pattern.  In  a  similar  manner,  a 
l>  of  pattern  is  struck  with  n  l>  of  profile  as  radius;  /'  b 
of  pattern  with/"  l>  of  Fig.  »>sl  as  radius;  //  <j'  of  pat- 
tern with  /'  <j"  of  Fig.  »>"!>  as  ratlins,  and y"  </'  of  pattern 
with /'  ij  of  prolilc  II  E  as  radius;  also,  b  c  of  pattern 
is  struck  with  b  c  of  profile  as  radius;  if  c  of  pattern 
with  #"  c  of  Fig.  681  as  radius ;  g'  h  of  pattern  with 
g  E  of  profile  as  radius,  and  c  h  of  pattern  with  c  h  of 


Fig.  tisn  as  radius.      Thus   arc   the   points    established 
in  O  II  -I    P  of  pattern. 

B  O  P  A  of  pattern  corresponds  with  B  O  P  A 
of  elevation  and  is  obtained  in  the  same  manner.  The 
points  in  0  B  of  pattern  are  derived  from  profile  L  B, 
as  are.  the  points  in  P  A  of  pattern  from  N  A  of  small 
profile.  The  lengths  of  solid  lines  in  pattern  are  ob- 
tained from  the  diagram  of  sections  in  Fig.  680,  as  are 
t-'k-ise  of  the  dotted  lines  from  the  diagram  of  sections 
in  Fig.  682.  Lines  drawn  through  the  points  in  B  O 
H  J  P  A,  Fig.  683,  will  be  the  half  pattern  for  ABO 
II  J  of  elevation.  The  other  half  of  pattern,  as  shown 
by  A  J'  II'  0'  B,  can  be  obtained  in  a  similar  manner 
or  by  duplication. 


PROBLEM  208. 


Pattern  for  a  Three-Pronged  Fork  With  Tapering;  Branches. 


In  Fig.  684  is  shown  a  pictorial  representation  of 
a  fork,  or  crotch,  consisting  of  three  branches  of  equal 
si/e  and  taper;  all  uniting  so  as  to  form  one  round 
pipe. 

In  the  plan,  Fig.  685,  ABC  represents  the  base 
of  article  or  size  of  the  large  pipe  and  B  D  E  C  G  one 
of  the  tapering  branches.  The  other  branches  are 
partly  shown  in  plan  by  A  (i  (1  S  T  and  A  II  V  B  G. 


Fill.  RS4. — I'l-rspri'tire  View  of  Three-Pronged  Fork  with  Tapering 
Branches. 

In  the  elevation  the  branch  is  shown  by  J  K  L  M  N 
and  the  half  profile  of  small  end  by  K  R  L. 

An  inspection  of  the  engraving  will  show  that  the 
perimeter  of  the  larger  end  of  the  branch  must  be  di- 
vided into  three  parts,  two  of  which  form  the  joints  or 
connections  with  the  branches  on  either  side  of  it 


while  the  third  part  must  form  one-third  of  the  base 
or  circumference  of  the  large  pipe  with  which  it  is  to 
be  united.  In  the  elevation  P  M  represents  the  plane 
of  the  base  or  upper  end  of  the  round  pipe  of  which 
A  B  C  is  the  profile  or  plan,  and  J  O  is  assumed  as 
the  hight  of  the  central  point  at  which  all  the  branches 
meet.  From  J  of  the  elevation  or  G  of  the  plan  to 
either  of  the  three  points  A,  B  or  C  any  suitable  curve 
may  be  chosen  as  the  profile  upon  which  to  make  a 
joint  or  miter  between  adjacent  branches.  As  J  0  is 
equal  to  G  A  or  G  C,  a  quarter  circle  is  assumed  as  the 
most  suitable  curve ;  therefore  from  O  as  a  center  de- 
scribe the  quarter  circle  P  J  of  the  elevation,  corre- 
sponding with  A  G  of  the  plan.  In  order  to  complete 
the  elevation  of  the  branch  J  K  L  M  N,  it  will  be 
necessary  to  obtain  the  elevation  of  the  miter  line  G  C. 
Therefore  divide  P  J  into  any  convenient  number 
of  equal  parts,  as  shown  by  the  small  figures,  and  from 
the  points  thus  obtained  carry  lines  to  the  right  paral- 
lel with  P  M.  From  G,  on  G  C,  set  off  spaces  equal 
to  the  distances  from  the  points  in  P  O  to  the  line 
0  J,  as  shown,  and  from  the  points  thus  obtained 
in  G  C  erect  perpendiculars  cutting  lines  of  similar 
number  drawn  from  P  J.  A  line  traced  through  these 
points  of  intersection,  as  shown  by  J  N,  will  give  the 
miter  line  in  elevation  corresponding  with  G  C  of  the 
plan.  Divide  C  H  of  the  plan  into  the  same  number 


392 


The  New  Metal    Worker  Pattern  Boole. 


of    equal   parts   as    P  J  of   the  elevation,  and  from  the 
points  thus  obtained  erect  perpendiculars  cutting  \  M. 
Divide   K  K  L,    the  profile   of  the   smaller  end  of  the 
branch,  into  the   same   number  of   equal    parts   as  the 
larger  end — that  is,  as  many  as  are  found  in  J  N  M— 
and  from   the   points   of  division    drop    lines    perpen- 
dicular to  K  L,  cutting  the  same.     Connect  points  in 
K  L  with  those  in  J  N  M  by  solid  and 
dotted  lines  in  the  manner    shown    in 
the  drawing.      Upon  all  of  these  lines 
it  will  be  necessary  to   construct  sec- 
tions in  order  to  obtain    the   true  dis- 
tances as  if  measured  upon  the  surface 
of   the  branch.      As  each  of  the  branch 
pipes  consists    of    symmetrical    halves 
when  divided  by  the  line  G  F  of  the 
plan    half  sections  only  need  be   con- 
structed, all  projections  being  measured 
from  the  dividing  plane  represented  1>\ 
Gr  F  in  the  plan  and  shown  in  eleva- 
tion by  J  K  L  M  N. 

In  Fig.    686  are  shown    the   sec- 
tions having  for  their  bases  the  solid 
lines  of  the  elevation,  which  are  con- 
structed   in     the    following    manner : 
I 'pon  any  horizontal- line,  as  P  Q,  set 
off  from  P  the  lengths  of  the  several 
solid  lines  of  the  elevation,  as  indicated 
by  the  small  figures  corresponding  with 
those  in  J  N  M.     At  P,  which  corre- 
sponds with   all  the  points  in  K  L   of 
the  elevation,  erect  a  perpendicular,  P 
H,  upon  which  set  off  the  hights  of  the 
points  in  K  R  L,  as  2'  2,   3'  3,  etc., 
shown  by  P  2,  P  3,  etc.     At  each   of 
the  points  near  Q  erect  a  perpendicu- 
lar, which  make  equal   in  hight  to  the 
length  of  line  drawn  from  the  point  of 
corresponding   number  in   G  C    II    of 
the  plan  to  the  line  G  II.     Thus  make. 
9'  9,   10'  10,  etc.,    equal   to  9"  a,  10" 
b,  etc.,  of  the  plan.     From  the  points  9, 
10,  etc.,  draw  solid  lines  to  the  points  in  H  P,  connect- 
ing points  correspondingly  connected  by  the  solid  lines 
of  the  elevation.     The  sections  having  for  their  bases 
the  dotted  lines  of  the  elevation  are  shown  in  Fig.  687, 
and    are    constructed    in    exactly    the    same    manner. 
Upon  Y  Z,  set  off  from  Y  the  lengths  of  the  dotted 
lines  of   the  elevation,   numbering  the  points  near  / 
to  correspond  with  those  in  J  N  M  of  the  elevation. 


The  perpendiculars   erected    from   these  points  are   tin- 
same   as   those   similarly    loraled    in    Fig.  (ISti.  and  the 
perpendicular  X  Y  is  a  duplicate  of  II  P  «(  Fig.  li.st;. 
From  the  points  9,  10,  12,  etc..  draw   dotted  lines  to    - 
points  in  X  Y,  connecting  points  correspondim'-lv  con-    *• 
nected  by  dotted  lines  of  the  elevation. 

To  describe  the  pattern  shown  in  Fig.    088  pro- 


1 


PROFILE 


ELEVATION 


/•>(/.  >;ft!),—Plan  and  Elevation  of  Three-Pronged  Fork. 

ceed  as  follows:  Draw  any  line,  as  .1  K,  in  length 
equal  to  J  K  of  elevation.  Fig.  685.  With  K  of  pat- 
tern as  center,  and  K  2  of  profile  as  radius,  describe  a 
small  arc  (2),  which  cut  with  one  struck  from  J  of 
pattern  as  center,  and  8'  2  of  Fig.  687  as  radius,  thus 
establishing  point  2  of  pattern.  With  point  2  of  pat- 
tern as  center,  and  9  2  of  Fig.  686  as  radius,  describe 
another  small  arc  (9),  which  intersect  with  one  struck 


Pattern  Problems. 


393 


from  J  of  pattern   as   center,  ant]   .1  !»'  of  elevation  ;i 
radius,  thus  establishing  the  point  9  of  pattern.     Pro- 


Fig.  686.— Diagram  nf  Sections  Upon  Solid  Lines  in  J  K  L  M  N 
of  Fig.  6Sr>. 

ceecl  in  this  manner  until  the  remaining  points  are 
located,  all  as  clearly  indicated  by  the  solid  and  dotted 
lines  in  Fig.  688.  By  drawing  lines  through  the 


""LIT  =~"~r4 

n»  — =•-=•— — 3  5 

_— _=«=.———— -— "  H3  6 
-=.— ^--=^--=~=~ """ I 


Ficj.  687.— Diagram  of  Sections  Upon  Dotted  Lines  in  J  K  L  M  N 
of  Firj.  686. 


points  thus  obtained  the  half  pattern  shown  by  K  Q 
L  M  N  J  is  the  result.     The  other  half,  as  shown  by 


Fiy.  688.—  Pattern  of  Taperinr/  Branch. 

K  Q'  L'  M'  N'  J,  can  be  obtained  in  a  similar  manner, 
or  by  duplication. 


PROBLEM  209. 


The  Pattern  for  an  Offset  to  Join  an  Oblon?  Pipe  With  a  Round  One. 


In  Fig.  689,  B  C  F  G  represents  the  side  elevation 
of  the  offset,  A  B  G  H  a  portion  of  a  round  pipe  join- 
ing it  below,  and  C  D  E  F  a  portion  of  the  oblong 
pipe  joining  it  above.  In  the  plan  immediately  below, 
•T  K  L  M  shows  the  plan  of  the  round  pipe  and  N  0  P 
Q  II  S  that  of  the  oblong  pipe,  while  the  distance. L  T 
shows  the  amount  of  the  offset. 

The  piece  forming  the  offset  is  similar  in  shape 
to  that  shown  in  Problem  189,  the  difference  being 
that  its  bases  B  G  and  C  F  are  neither  horizontal  nor 
parallel  to  each  other  and  that  sections  on  the  lines  of 
the  bases  are  not  given.  Since  the  article  required 
consists  of  symmetrical  halves  when  divided  on  the 
line  J  T  of  the  plan,  the  plane  surface  ABCDEFGH 


lying  as  it  were  back  of  the  half  shown  by  the  eleva- 
tion may,  as  in  Problem  207,  be  regarded  as  a  base 
from  which  to  measure  all  hights,  or  projections,  in 
obtaining  the  required  profiles  and  sections  necessary 
in  developing  the  pattern.  The  first  steps  necessary 
will  be  to  obtain  true  sections  upon  the  lines  C  F 
and  B  G  of  the  elevation.  In  Fig.  690,  C  D  E  F  rep- 
resents a  duplicate  of  the  part  bearing  the  same  letters 
in  the  elevation.  Upon  D  E  as  a  base  line  construct 
a  duplicate  of  the  half  section  of  oblong  pipe  N  0  P  T 
of  Fig.  689,  as  shown  by  D  N  P  E. 

Divide  the  semicircle  N  P  into  any  convenient 
number  of  equal  parts,  as  shown  by  the  small  figures. 
With  the  blade  of  the  T-square  placed  at  right  angles 


The   Xi-ir    Mi-Utl     \VorL-iT    I'ntU-riL 


to  1)  E,  drop  lines  cutting  C  F.  "With  the  T-s1uarc 
placed  at  right  angles  to  C  F,  mid  brought  Against  the 
points  in  C  F,  draw  lines,  extending  them  indefinitely, 
as  shown.  Measuring  in  each  instance  from  C  F,  set 
off  on  the  lines  just  drawn  the  same  length  as  similar 
lines  in  D  N  O  P  E,  and  through  the  points  thus  ob- 


M 


Fit/.  689.— Plan  and  Elrrntinn  nf  Offset. 

tained  trace  a  line,  as  shown  by  n  0  p.  Then  C  F  p 
O  n  is  the  half  shape  of  cut  on  line  C  F.  In  Fig.  691, 
A  B  G  II  is  a  duplicate  of  the  elevation  of  the  round 
pipe,  below  which  is  drawn  a  half  profile  of  same,  A 
M  II.  To  obtain  the  shape  of  cut  on  line  B  G,  divide 
the  half  profile  A  M  H  into  the  same  number  of  parts 
as  was  N  P,  and,  with  the  T-square  placed  parallel 


with  A  1>,  and  brought  successively  against  the  points 
in  A  M  II,  carry  lines  cutting  ]><!.  With  the  T-square 
placed  at  right  angles  to  B  G,  and  brought  against  the 


Fig.  690. — Development  of  Section  on  Line  C  F  of  Elevation. 

points  therein  contained,  erect  perpendiculars,  as  shown. 
Measuring  in  each  instance  from  B  G,  set  off  on  the 
lines  just  drawn  the  same  length  as  similar  lines  in  A 


14  H 


Fill.  091. — Development  of  Section  on  Line  B  G  of  Elevation. 

f 

M  H,  and  through  the  points  thus  obtained  trace  a  line, 
as  shown  by  B  m  G.  Then  B  m  G  is  the  half  shape 
of  cut  on  line  B  G. 

In  order  to  avoid  a  confusion  of  lines  a  duplicate 


Pattern  Problems. 


395 


of  15  C  V  <i  of  the  elevation  is  presented  in  Fig.  »>'.»•_', 
upon  C  V  and  B  G  of  which,  as  base  lines,  are  drawn 
duplicates  of  the  sections  obtained  in  Figs..  <>!»0  and 
1191,  all  as  shown.  From  points  in  u  0  p  drop  lines  at 
right  angle's  to  C  !•'.  cutting  the  same,  and  from  points 
in  B  in  (1  drop  lines  at  right  angles  to  B  G,  cutting  it. 
Connect  points  in  these  lines  iu  consecutive  order  by 
solid  lines,  as  shown,  and  subdivide  the  four  sided 
figures  thus  obtained  by  dotted  lines  representing  their 
shorter  diagonals.  The  surface  of  the  offset  or  transi- 
tion piece  is  thus  divided  into  a  series  of  very  tapering 
triangles,  the  lengths  of  whose  bases  or  shortest  sides 


Fiij.  «».'.—  Middle  Piece  of  Offset,  Showiny  Method  of  Triangulation. 

are  given  in  the  two  sections  C  n  0  p  F  and  B  m  G. 
In  order  to  obtain  the  correct  lengths  of  their  longer 
sides  two  diagrams  or  series  of  sections  must  be  con- 
structed for  that  purpose,  which  arc  shown  in  Figs. 
093  and  <i'.»4. 

To  obtain  the  various  sections  on  the  solid  lines 
of  the  elevation  proceed  as  follows:  Draw  the  right 
angle  V  V  W,  Fig.  693.  From  V,  on  V  U,  set  off 
the  length  of  lines  in  B  m  G,  Fig.  692.  From  V,  on 
V  W,  set  off  the  length  of  solid  lines  in  B  C  F  G,  and 
from  the  points  thus  obtained  erect  perpendiculars,  in 
length  equal  to  lines  of  similar  number  in  0  n  0  p  F, 
l-'ig.  692.  Thus  make  line  W  1'  of  Fig.  693  equal  to 


line  C  n  or  C  1,  and  draw  V  1',  which  gives  the 
distance  from  point  B  to  point  n  in  Fig.  692  as  if 
measured  on  the  finished  article.  Connect  the  perpen- 
diculars drawn  from  V  W  with  the  points  in  V  U,  as 
shown,  and  corresponding  with  the  figures  in  Fig.  C92. 
Thus  connect  1'  and  8,  2'  and  9,  3'  and  10,  etc.  Then 


769     4321V/ 


Fly.  ti'JJ.— Diagram  of  Sections  on  Solid  Lines  of  Fiy.  692. 

will  the  lengths  of  the  oblique  lines  in  Fig.  693  be  the 
true  lengths  of  the  solid  lines  crossing  the  elevation. 
The  diagram  of  sections  shown  in  Fig.  694  is  con- 
structed in  the  same  manner,  using  the  dotted  lines  of 
the  elevation  as  the  basis  of  measurements. 

Draw  the  right  angle  X  Y  Z,  and  from  Y  set  off 
on  Y  Z  the  length  of  lines  in  C  n  0  p  F.  From  Y,  on 
Y  X,  set  off  the  length  of  dotted  lines  in  B  C  F  G,  and 
from  the  points  thus  obtained  erect  perpendiculars,  in 
length  equal  to  lines  of  similar  number  in  B  m  G,  as 
indicated  by  the  small  figures.  Connect  the  perpen- 
diculars drawn  from  X  Y  with  the  points  in  Y  Z,  as 
shown,  and  corresponding  with  the  figures  in  B  C  F  G. 
Thus  connect  1  and  9',  2  and  10',  3  and  11',  etc. 


Fiy.  694. — Diagram  of  Sections  on  Dotted  Lines  of  Fiy.  692. 

An  inspection  of  the  plan  and  elevation,  Fig.  689, 
will  show  that  the  curved  surface  of  the  offset  or  transi- 
tion piece  B  C  F  G,  which  has  been  divided  into  tri- 
angles, is  shown  by  J  M  L  Q  R  S  of  the  plan,  and  that 
this  piece  is  connected  with  its  mate  or  equivalent  in 


396 


Tli 


Metal    \\'vrk<:r    Pattern    />W  . 


the  opposite  lialf  of  the  article  by  ;i  large  plain  trian- 
gular surface,  S  J  N,  on  the  upper  side,  and  by  another, 
Q  L  P,  on  its  lower  side,  which  must  be  added  to  the 
pattern  of  the  curved  portion  after  it  has  been  de- 
veloped. It  will  also  lie  seen  that  V  W  1'  of  Fig. 
093  is  one-half  of  J  N  S.  Therefore'  to  develop  the 
pattern,  first  draw  any  line,  as  j  x  in  Fig.  (595,  in  length 
equal  to  B  C  of  Fig.  689,  or  V  W  of  Fig.  693.  With 
/  as  center,  and  1'  V  of  Fig.  693  as  radius,  describe  a 
small  arc  (near  ,v),  which  intersect  with  another  small 


intersect  with  another  .small  arc  struck  from  point  1  of 
pattern  as  center,  and  a  radius  equal  to  1  2  of  the  pro- 
file ('  n  ()  ji  F  of  Fig.  692,  thus  establishing  the  position 
of  point  2  of  pattern.  Proceed  in  this  manner,  using  the 
dotted  oblique  lines  in  Fig.  t>;»4,  the  lengths  of  the 
spaces  in  B  in  G  in  I^ig.  692,  the  lengths  of  the  solid 
oblique  lines  in  Fig.  693  and  the  lengths  of  the  spaces 
in  0  n  O  p  F  of  Fig.  692  in  the  order  named  until  the 
line  7  14  is  reached.  Lines  traced  through  the  points 
of  intersection  fromj  to  I  and  from  s  to  t  will  give  the 


Fig.  69't.— Pattern  of  Offset. 


arc  struck  from  x  as  center,  and  with  a  radius  equal  to 
W  1'  of  Fig.  693,  thus  duplicating  the  triangle  V  W 
1'.  From  s,  or  point  1,  as  center,  with  a  radius  equal 
to  1  9'  of  Fig.  694,  describe  a  small  arc  (near  9),  which 
intersect  with  another  small  arc  struck  from  j  or  8  of 
pattern  as  center,  with  a  radius  equal  to  8  9  of  the  pro- 
file B  m  G,  Fig.  692,  thus  establishing  the  position  of 
point  9  of  pattern.  From  9  as  a  center,  with  a  radius 
equal  to  9  2'  of  Fig.  693,  describe  a  small  arc,  which 


shape  of  the  curved  portion,  of  the  pattern.  From  14 
of  pattern  as  center,  with  a  radius  equal  to  G  F  of  Fig. 
I'.'.H.  or  V  7  of  Fig.  693,  describe  a  small  are,  which 
intersect  with  another  small  arc  struck  from  point  7  of 
pattern  as  center,  and  a  radius  equal  to  7  7' of  Fig. 
•  '•It::,  or  T  P  of  the  plan.  Draw  7  t  and  t  /;  then  will 
/_/  .':  .s  rj  t  be  one-half  the  pattern  required.  The  other 
half  can  be  obtained  by  any  means  of  duplication 
most  convenient. 


PROBLEM    210. 

Pattern  for  an  Offset  to  Join  a  Round  Pipe  with  one  of  Elliptical  Profile. 


This  problem  differs  from  the  preceding  one  only 
in  the  shape  of  the  pipe  having  the  elongated  profile, 
which  profde  in  the  preceding  problem  consists  of  two 


semicircles  joined  by  a  straight  part,  whereas  in  this 
ease  its  curve  is  continuous  throughout;  its  pattern 
therefore  will  consist  throughout  of  a  series  of  triangles 


Pattern  Problems. 


having  short  buses  instead  of  having  a  large  llat  trian- 
gular surface  uniting  its  curved  portion  as  in  the  pre- 
vious ease. 

Ill  Fig.  li'.Mi.  1)  ('  1!  A  represents  the  elevation  of 
the  otl'set.  ('  F  K  15  that  of  a  portion  <>f  the  round  pipe 
with  which  it  is  required  to  connect  at  its  upper  end 


?'; 

r7!? 

-& 

i      I 

' 

i      i 

; 

i      i 
i      | 

ELEVATIOM 


\ 


\ 


PROF 


Fiij.  G96.— Elevation  and  Sections  of  Offset,  Shonriny  Method  of 
Triangulation. 


and  II  1)  A.  U  that  <>f  the  elliptical  pipe  joining  it  be- 
low. M  I'  X  is  the  half  profile  of  the  round  pipe  and 
K  J  L  that  of  the  elliptical  pipe.  The  plan,  or  top 
view  is  not  shown,'  and  is  not  necessary  to  the  work  of 
obtaining  the  pattern.  Since  the  profiles  given  neces- 
sarily represent  sections  on  lines  at  right  angles  to  the 
respective  pipes,  as  at  I1'  K  and  II  0,  it  will  .first  be 
necessary  to  derive  from  them  sections  on  the  joint  or 


miter  lines  C  1>  and  1>  A,  from  which  to  obtain  correct 
stretchouts  of  the  two  ends  of  the  pattern  of  the  offset 
piece. 

As  the  pattern  required  consists  of  svinmet- 
rical  halves,  one-half  only  will  be  given,  and  one-half 
of  the  profiles  only  need  be  used.  Therefore  divide 
the  half  profile  M  1'  X  into  any  convenient  number  of 
equal  spaces,  as  shown  by  the  small  figures,  and  from 
the  points  thus  obtained  draw  lines  at  right  angles  to 
F  E,  cutting  M  N  and  C  13.  To  avoid  confusion  of 
lines  a  duplicate  of  C  B  is  shown  at  the  left  by  C1  B1. 
From  the  points  on  C'  B'  draw  lines  at  right  angles  to 
it  indefinitely,  and  upon  each  of  these  lines,  measuring 
from  C'  B',  set  off  the  lengths  of  lines  of  corresponding 
number  in  the  profile  M  P  N  measured  from  M  N.  Thus 
make  the  distance  of  point  2'  from  C'  B'  equal  to  the 
distance  of  point  2  from  line'M  N,  the  length  of  line 
3'  equal  to  that  of  line  3,  measuring  from  the  same  base 
lines  as  before,  etc.  A  line  traced  through  the  points 
of  intersection,  as  shown  by  C1  0  B1,  will  be  the  correct 
section  on  the  line  C  B  of  the  elevation.  The  method 
of  obtaining  the  section  on  the  line  D  A,  shown  at  D' 
I  A',  is  exactly  the  same  as  that  just  described  in  con- 
nection with  the  round  pipe,  all  as  clearly  shown  in  the 
lower  part  of  the  engraving. 

The  next  operation  will  consist  of  dividing  the 
surface  of  the  transition  or  offset  piece  into  measurable 
triangles,  making  use  of  the  spaces  used  in  the  profiles; 
therefore  connect  points  in  C  B  with  those  of  similar 
number  in  D  A  by  solid  lines,  as  1  with  1',  2  with  2', 
.etc.,  and  connect  points  in  C  B  with  those  of  the  next 
higher  number  in  D  A  by  dotted  lines,  as  1  with  2',  2 
with  3',  3  with  4',  etc.  The  surface  of  the  transition 
piece  is  thus  divided  into  a  series  of  triangles  the 
lengths  of  whose  bases  or  short  sides  are  found  in  the 
two  sections  C'  0  B1  and  D1  I  A1. 

As  the  bights  of  corresponding  points  in  the  two 
sections,  measuring  from  their  center  or  base  lines, 
differ  very  materially,  it  will  be  necessary  to  construct 
two  diagrams  of  sections  from  which  the  lengths  of  the 
various  solid  and  dotted  lines  can  be  obtained.  In 
Fig.  697  is  shown  a  diagram  of  sections  through  A  B 
C  D  taken  on  the  solid  lines  drawn  across  the  eleva- 
tion, in  which  the  base  line  P  Q  represents  the  sur- 
face of  a  plane  dividing  the  offset  into  symmetrical 
halves.  At  P  erect  a  perpendicular,  P  E,  upon  which 
set  off  the  hight  of  the  points  in  the  profile  K  J  L  or 
the  section  I)'  I  A1,  measuring  upon  the  straight  lines 
joining  them  with  the  base  line  K  L,  as  shown  by  the 
small  figures.  From  P,  upon  P  (j,  set  off  the  lengths 


398 


Xvw  Metal    \YUI-JM-  Pattern  Bwk. 


of  the  various  solid  lines  drawn  across  the  elevation, 
also  shown  by  small  figures,  and  at  each  of  the  points 
thus  obtained  erect  a  perpendicular,  which  make  equal 
in  hight  to  the  distance  of  point  of  corresponding  num- 
ber in  profile  M  P  N  from  M  N,  measuring  on  the  per- 
pendicular line.  Thus,  make  line  2  of  Fig.  697  equal 


IB 


f'iy.  <J'J7.—Dia<jram  of  Sections  on  Solid  Lines  of  Elevation. 

in  hight  to  the  distance  from  point  2  of  profile  to  the 
line  M  N,  line  3  equal  to  the  length  of  line  3  of  profile 
M  P  N.  Now  connect  points  of  corresponding  num- 
ber at  the  two  ends  of  the  diagram  by  straight  lines, 
as  shown,  then  will  these  oblique  lines  be  the  correct 
distances  between  points  of  corresponding  numbers 
connected  by  the  solid  lines  drawn  in  the  elevation. 

The  diagram  in  Fig.  698  is  constructed  in  an  ex- 
actly similar  manner.  The  distances  S  1,  S  2,  S  3, 
etc.,  on  the  base  line  are  in  this  case  made  equal  to  the 
lengths  of  the  dotted  lines  of  the  elevation,  and  the 


,U 


87 


28 


ID 


Fig.  698. — Diagram  of  Sections  on  Dotted  Lines  of  Elevation. 

perpendiculars  erected  at  points  2,  3,  etc.,  are  the 
same  as  those  used  in  the  previous  diagram.  The  per- 
pendicular S  U  is  also  an  exact  duplicate  of  P  R  in 
Fig.  697.  In  drawing  the  oblique  dotted  lines,  point 
1  at  the  right  end  of  the  diagram  is  connected  with 
that  of  the  next  higher  number  (2')  on  the  line  S  U,  2 
at  the  right  with  3'  on  the  line  S  U,  etc.,  all  as  shown. 


The  oblique  dotted  lines  will  then  be  the  correct  dis- 
tances between  points  of  corresponding  numbers  con- 
nected by  the  dotted  lines  in  the  elevation. 

To  develop  the  pattern,  first  draw  any  straight 
line,  as  1)  C  in  Fig.  699,  which  make  equal  in  length  to 
D  C  of  Fig.  696.  From  1)  as  center,  with  a  radius 
equal  to  1  -2  of  the  section  1)'  I  A',  strike  a  small  arc, 
which  intersect  with  another  small  arc  struck  from  (_' 
as  center,  with  a  radius  equal  to  2'  1  of  Fig.  6)is,  thus 
establishing  the  location  of  point  '!'  of  pattern.  From 
•1'  of  pattern  as  center,  with  a  radius  equal  to  •>.'  '2  "I 
Fig.  697,  strike  a  small  are,  which  intersect  with  an- 


Fig.  690.— Half  Pattern  of  Offset  Piece. 

other  small  are  struck  from  0  of  pattern  as  center,  and 
a  radius  equal  to  1  2  of  section  C1  0  B',  thus  establish- 
ing the  position  of  point  2  of  pattern.  So  continue. 
using  alternately  the  dotted  and  the  solid  oblique  lines 
in  Figs.  698  and  697  to  measure  the  distances  across 
the  pattern,  the  spaces  from  the  section  1)'  1  A1  to 
form  the  stretchout  of  the  lower  end  (I)  A)  of  pattern, 
and  the  spaces  from  the  section  C'  O  B1  to  form  tin- 
stretchout  of  the  upper  end  (C  B)  of  the  pattern.  Lines 
traced  through  the  points  of  intersection,  as  from  C  .to 
B  and  from  I)  to  A,  will  complete  one-half  the  required 
pattern. 


Pattern  Problems.  399 

PROBLEM   211. 
The  Patterns  for  a  Funnel  Coal  Hod. 

In  Fig.  Too   HIV  shown  the  drawings  for  a  funnel  IK-IT:  constructed  in  two  pieces,  the  front  being  in  one 

coal  hod  of  a  style  in  general  use.      In  preparing  such  piece  joined  together  on  the  line  B  C  of  the  elevation 

a   set   of  drawings   it  is  necessary  that  care  should  lie  or  B3  C3  of   the  plan,  and   joined   to   the  hack  piece  on 

taken   to   have    a  correspondence  of  all  the  principal  the  line  II  D.      As  will  be   seen    by  an  inspection  of 


c' 


Fill.  7'iJO—riiin,   KIrration  inul  Xfrtions  of  a  Funnel  Coal  Hod,  Shoiriiifi  Method  of  Trian  (filiation. 


parts  in  the  two  views,  as  shown  by  the  dotted  lines, 
leaving  the  final  drawing  of  the  curves  to  be  more 
accurately  performed  as  circumstances  may  require  in 
subsequent  parts  of  the'  work.  The  design  is  capable 
of  any  degree  of  modification  so  far  as  the  proportions 
of  its  parts  are  concerned  without  in  the  least  affect- 
ing the  method  of  obtaining  its  patterns.  Thus, 
hights,  lengths,  diameters  or  curves  may  be  changed 
at  the  discretion  of  the  designer.  The  coal  hod  is 


the  elevation,  the  front  piece  consists  of  a  flat  tri- 
angular piece.  II  .1  D,  joined  to  two  irregular  flaring 
pieces,  A  .1  II  (i  and  15  J  D  C.  On  account  of  the 
taper  or  slant  of  the  flat  portion  of  the  front  piece,  as 
shown  by  J2  I)2  of  the  plan,  the  line  D2  H'  has  been 
drawn  somewhat  obliquely  from  X-,  the  center  of  the 
bottom,  instead  of  at  right  angles  to  A2  E". 

The   section  at   A  B    is  assumed  to  be  a  perfect 
circle   and    should    be    drawn    exactly    opposite,    as 


400 


The  New  Metal    Worker  i'attcru  Book. 


shown,  its  vertical  center  line  A1  B1  being  placed 
parallel  to  A  B.  Divide  each  quarter  of  this,  as  A1  J' 
and  J1  B',  into  any  number  of  equal  spaces,  as  shown 
by  the  small  figures,  and  through  the  points  thus  ob- 


Q  87'   fl'     5 

Fig.  701.— Diagram  of  Sections  on  Solid  Lines  in  A  J  H  G  of  Fig.  700. 

tained  draw  lines  cutting  A1  B1  and  A  B.  From  J, 
the  middle  point  on  A  B,  draw  lines  to  D  and  to  H. 
Also  divide  H'  G1  of  the  plan  into  the  same  number 
of  equal  spaces  as  A1  J1,  numbering  the  points  to  cor- 
respond. From  the  points  thus  obtained  erect  lines 
perpendicularly,  cutting  G  H  of  the  elevation.  Con- 
nect points  ot  like  number  on  A  J  and  G  H,  as  5  with 
5,  6  with  6,  etc.,  by  solid  lines,  as  shown;  also,  con- 
nect each  point  on  A  J  with  that  of  next  higher 
number  on  G  H  by  a  dotted  line,  as  5  with  6,  6  with 
7,  etc.  These  solid  and  dotted  lines  just  drawn  are 
the  lines  upon  which  measurements  are  to  be  taken  in 
obtaining  the  pattern,  and  upon  which  sections  must 


ftl 


- 

______  _  ~"    ^_"~r"~-  -  --^1-     J 


,U 


T  98'   7'  6 

Fig.  702.— Diagram  of  Sections  on  Dotted  Lines  in  AJHGof  Fig.  700. 

be  constructed  before  their  true  lengths  can  be  ob- 
tained. 

In  Fig.  701  are  shown  the  sections  having  the 
solid  lines  in  A  J  H  G  as  their  bases,  which  are  con- 
structed in  the  following  manner:  Draw  any  right 
angle,  as  P  Q  K.  Upon  P  Q  set  off  the  hights  of  the 
several  points  in  the  section  A1  J'  from  the  line  A1  B1, 
as  measured  upon  the  straight  lines  joining  them  with 
A'  B1 ;  thus  make  Q  5  and  Q  6  equal  to  the  distance 
of  points  5  and  6  from  the  line  A1  B1.  From  Q  on 
Q  B,  measuring  from  Q,  set  off  the  lengths  of  the 
several  solid  lines  in  A  J  H  P,  as  indicated  by  the 
small  figures,  and  from  the  points  thus  obtained  erect 
perpendiculars  equal  in  hight  to  the  length  of  lines 
drawn  from  points  of  corresponding  number  in  G'  11' 
of  the  plan  to  the  line  G1  X;  thus  make  the  perpen- 


diculars at  points  5',  6',  etc.,  equal  to  the  length  of 
the  lines  drawn  from  points  5  and  6  in  G'  II'  to  G1  X. 
Connect  the  points  thus  obtained  with  points  of  cor- 
responding number  in  P  Q.  The  oblique  lines  thus 
obtained  will  be  the  true  distances  represented  bv 
lines  of  corresponding  number  in  the  elevation.  The 
diagram  in  Fig.  7(>2  shows  the  sections  upon  the  dotted 
lines  in  A  J  II  G  and  is  constructed  in  the  same  man- 
ner. Upon  T  U,  .measuring  from  T,  are  set  off  the 
lengths  of  the  several  dotted  lines.  S  T  is  the  same 
as  P  Q  of  Fig.  701,  and  the  perpendiculars  at  U  are 
equal  to  those  of  corresponding  number  in  Fig.  701. 
Points  in  S  T  are  then  connected  with  the  perpen- 
diculars of  next  higher  number  by  dotted  lines,  which 


L  3'3'    4'         5 

Fig.  70S.— Diagram  of  Sections  on  Solid  Lines  in  J  B  C  D  of  Fiij.  7<i<>. 

give  the  true  lengths  represented  by  the  dotted  lines  of 
the  elevation. 

That  portion  of  the  front  piece  shown  by  .1  B  ('  1> 
of  the  elevation  must  be  triangulated  in  exactly  the 
same  manner  as  the  portion  just  described,  and  sec- 
tions constructed  upon  the  several  solid  and  dotted 
lines  there  drawn,  as  shown  in  Figs.  703  and  704. 
However,  as  no  outline  is  given  in  either  the  plan  or 
the  elevation  from  which  a  correct  stretchout  of  C  1> 
can  be  obtained,  a  section  must  be  constructed  for  that 
purpose,  which  can  be  done  in  the  following  manner : 
First  draw  C1  M1  as  the  vertical  center  line  of  a  rear 


;r -i.i.  I.- 

-,W  !"  3      4 


-.w 

Fig.  704.— Diagram  of  Sections  on  Dotted  Lines  in  J  B  C  D  of  Fig.  700. 

elevation.     From  points  C  and  D  project  lines  hori- 
zontally to  the  ri^'lit,  cutting'  C'  M1  at  C'  and  M.     l'| 

D  M,  measuring  from  M,  set  off  half  the  widtli  of  the 
front  piece  at  D"  of  the   plan;   that  is,   make  M  D1 


Pattern  Problems. 


401 


equal  to  M"  T>\  Any  desirable  curve  may  then  be 
drawn  from  1)'  to  C',  representing  the  rear  elevation  of 
curve  represented  by  D  0  of  the  side  elevation.  As 
the  distance  from  C  to  D  is  much  greater  than  C'  M, 
an  extended  profile,  as  measured  upon  C  D,  must  n<>\v 
lie  developed  from  which  to  obtain  a  correct  stretch- 
out of  that  portion  of  the  pattern. 

Therefore  divide  the  curve  C'  D1  into  the  same 
number  of  parts  as  the  quarter  circle  B'  J',  and  from 
the  points  thus  obtained  carry  lines  horizontally  to  the 
left,  cutting  C  D.  Upon  C1  M  extended,  as  C*  M1,  set 
off  spaces  equal  to  those  in  C  D,  as  shown,  and  through 


Fig.  705.—  Half  Pattern  of  Front  Piece  of  Funnel  Coal  Hod. 


the  points  thus  obtained  draw  lines  to  the  left  indef- 
initely. From  the  points  in  C1  D1  drop  lines  verti- 
cally, cutting  those  just  drawn,  all  as  shown.  A  line 
traced  through  the  points  of  intersection,  as  shown 
from  D"  to  C2,  will  give  the  desired  stretchout.  In 
dividing  the  curve  C'  D1  into  spaces  it  is  advisable  to 
make  those  nearest  to  D1  less  than  those  near  the  top 
of  the  curve  in  order  to  compensate  for  the  increase  in 
the  spaces  in  C"  M'  as  they  approach  the  bottom; 
thus  obtaining  a  set  of  nearly  equal  spaces  upon  the 
final  profile  D2  C2,  all  of  which  will  appear  clear  by  an 
inspection  of  the  drawing, 

As  above  stated,  the  diagrams  of  sections  in  Figs. 
7<>3  and  704  are  constructed  in  the  same  manner  as 


those  of  Figs.  701  and  702.  The  bights  in  K  L  and 
V  W  are  taken  from  J1  B'  of  Fig.  700  and  are  the 
same  as  those  in  P  Q  and  S  T  of  Figs.  701  and  702. 
The  distances  upon  L  N  and  W  Y  are  those  of  the 
solid  and  dotted  lines  in  J  B  C  D  of  Fig.  700,  and  the 
hights  of  the  perpendiculars  near  N  and  Y  are  equal 
to  the  lengths  of  the  lines  drawn  from  points  of  cor- 
responding number  in  the  profile  D2  C"  of  Fig.  700  to 
the  lines  C'  M1. 

To  develop  the  pattern  of  the  front  piece,  first 
draw  any  line,  as  A  G  in  Fig.  705,  equal  in  length  to 
A  G  of  Fig.  700.  From  G  as  a  center,  with  a  radius 
equal  to  the  dotted  lines  9  8  of  Fig.  702,  describe  a 
short  arc  (near  8),  which  intersect  with  another  arc 
drawn  from  A  as  center,  with  a  radius  equal  to  9  8  of 
the  section  A1  J'  B1  of  Fig.  700,  thus  establishing  the 
position  of  point  8  in  the  upper  line  of  the  pattern. 
From  8  of  the  pattern  as  center,  with  a  radius  equal 
to  8  8  of  Fig.  701,  describe  a  short  arc  (near  8'),  which 
intersect  with  another  arc  drawn  from  G  of  the  pat- 
tern as  center,  with  a  radius  equal  to  9  8  of  the  plan, 
Fig.  700,  thus  establishing  the  point  8'  in  the  lower 
line  of  the  pattern.  Continue  in  this  manner,  using 
the  lengths  of  the  oblique  dotted  lines  in  Fig.  702  in 
connection  with  the  spaces  in  the  section  A1  J1  B1  of 
Fig.  700  as  radii  to  determine  the  points  in  the  upper 
line  of  the  pattern,  or  the  side  forming  the  mouth,  and 
the  lengths  of  the  oblique  solid  lines  of  Fig.  701  in 
connection  with  the  spaces  in  the  plan  of  the  bottom 
(G1  H')  as  radii  with  which  to  determine  the  points  in 
the  lower  line  of  the  pattern  or  the  side  to  fit  against 
the  bottom. 

Having  reached  the  points  5  and  5',  next  add  to 
the  pattern  the  flat  triangular  surface  shown  by  J  H  D 
of  the  elevation.  From  H  (5')  of  the  pattern  as  center, 
with  a  radius  equal  to  5  5  of  Fig.  706,  the  side  of  the 
last  triangle  in  the  pattern  of  the  back  piece,  describe 
a  short  arc  (near  D),  and  intersect  the  same  with 
another  are  struck  from  J  (5)  of  the  pattern  as  center, 
with  a  radius  equal  to  the  oblique  line  5  5  of  Fig.  703, 
and  draw  H  D  and  B  J.  Using  D  J  of  the  pattern  as 
one  side  of  the  next  triangle,  take  as  radii  the  dis- 
tances 5  4  of  Fig.  704  and  5  4  of  the  section  D2  C1  of 
Fig.  700  to  locate  the  position  of  point  4'  of  the  pat- 
tern, as  shown  in  Fig.  705.  With  4  4  of  Fig.  703, 
and  5  4  of  the  section  B'  J'  of  Fig.  700  as  radii  locate 
the  point  4  of  the  pattern,  as  shown,  and  so  continue 
until  C  D  is  reached.  Lines  traced  through  the  points 
of  intersection  from  B  to  A,  C  to  D  and  H  to  G  will 
complete  the  pattern  of  one-half  the  front  piece. 


402 


Tlie  New  Metal    \Vorkei-  1'uttcni 


The  method  of  triangulating  the  piece  forming 
tlie  back  of  the  coal  hod  and  the  development  of  the 
pattern  of  the  same  are  so  clearly  shown  in  Figs. 
706,  707  and  70S,  in  addition  to  the  plan  and  elevation, 
Fig.  700,  as  to  need  only  a  brief  description.  Divide 
H'  F'  and  Da  E2  of  the  plan,  Fig.  700,  into  the  same 


Fig.  706.— Diagram  of  Sections  on  Solid  Lines  in  D  E  F  H  of  Fig.  700. 


number  of  equal  parts,  and  from  the  points  thus  ob- 
tained erect  lines  vertically  cutting  the  corresponding- 
lines  II  F  and  D  E  of  the  elevation,  as  shown  by  the 
dotted  lines.  Connect  points  of  like  number  in  that 
view  by  solid  lines  and  points  in  D  E  with  those  of 
next  lower  number  in  II  F  by  dotted  lines.  Since  D  E, 
being  inclined,  is  longer  than  M2  E3,  its  equivalent 
in  the  plan,  it  will  be  necessary  to  develop  an  ex- 
tended section  upon  the  line  D  E  of  the  elevation,  as 
shown  by  D5  E3  of  the  plan,  which  may  be  done  in 
the  same  manner  as  the  section  on  the  line  C  D  above 
explained.  Upon  Ma  Ea  extended,  as  E"  E',  set  off  the 


Fig.  707.— Diagram  of  Sections  on  Dotted  Lines  in  D  EF  H  of  Fig.  700. 


spaces  in  D  E,  and  through  the  points  thus  obtained 
draw  lines  at  right  angles,  as  shown,  which  intersect 
with  lines  drawn  parallel  with  M*  E"  from  points  of 
corresponding  number  in  D"  E",  thus  establishing  the 
curve  D6  E3,  from  which  a  correct  stretchout  of  the  top 
of  the  back  piece  may  be  obtained. 


In  Figs.  7o(i  and  7<>7,  the  hights  of  the  various 
points  upon  the  perpendiculars  from  X  and  Z  are  equal 
to  the  lengths  of  the  straight  lines  drawn  from  points 
of  corresponding  numbrr  in  II'  F1  of  the  plan,  Fig. 
700,  to  the  line  MJ  F1.  The  distances  set  off  to  tin- 
right  upon  the  horizontal  lines  from  X  and  Z  are  equal 
to  the  lengths  of  the  several  solid  and  dotted  lines  ir> 
D  E  F  II  of  the  elevation,  and  the  hights  of  the  per- 
pendiculars at  the  right  ends  of  the  bases  are  equal  to 
the  straight  lines  drawn  from  points  in  the  section 
Ds  E3  to  the  line  E"  E3.  The  several  oblique  solid  and 
dotted  lines  are,  therefore,  the  true  distances  repre- 
sented by  the  solid  and  dotted  lines  of  corresponding 
number  in  the  elevation. 


Fig.  708.— Half  Pattern  of  Back  Piece  of  Funnel  Coal  Hud. 


In  Fig.  708,  E  F  is  equal  to  E  F  of -Fig.  700  and 
is  made  the  base  of  the  first  triangle,  from  which  base 
the  several  triangles  constituting  the  complete  pattern 
may  be  developed  in  numerical  order  and  in  the  usual 
manner  from  the  dimensions  obtained  in  Figs.  706  and 
707  and  in  the  plan  and  section  in  Fig.  700,  all  as 
clearly  indicated. 

The  pattern  for  the  piece  forming  the  foot  of  the 
coal  hod  is  a  simple  frustum  of  an  elliptical  cone,  the 
method  of  obtaining  which  is  fully  explained  in  Prob- 
lem 171.  In  Fig.  546  of  that  problem  the  lines  E  F  ami 
Gr  H  are  drawn  much  further  apart  than  the  proportions 
of  the  foot  in  the  present  case  would  justify,  but  the 
operation  of  obtaining  its  pattern  is  exactly  the  same. 


Pattern  Problems. 

PROBLEM   212. 


4u:; 


Patterns  for  a  Three-Piece  Elbow  to  Join  a  Round  Pipe  with  an  Elliptical  Pipe. 


In  Fig.  709,  let  A  B  C  D  represent  the  profile  of 
the  round  pipe  and  EFGHIJKL  the  elevation  of 
the  elbow.  In  the  plan  the  profile  of  round  pipe  is 
represented  l>y  A1  B1  C"  D1,  the  elbow  by  Pl  M'  G1  M' 


PROFILE 
C 


ELEVATION 


Fig.  709.— Plan  and  Elevation  of  Three-Piece  Elbow,  Round  at  One  End  and 
Elliptical  at  the  Other. 


P",  and  the  shape  of  elliptical  end  of  elbow  by  J1  M1 
G1  M".  The  section  J  G  II  I  of  elevation  is  without 
Hare,  and  sections  K  F  G  J  and  E  F  K  L  are  flared,  as 
shown  in  plan.  Through  the  plan  draw  E1  G1  and  M1 
M",  and  carry  M'  M'1  through  the  center  of  elevation, 
as  shown  by  M  N  ()  P.  Perpendiculars  dropped  from 
the  points  K  O  F  of  elevation,  cutting  P'  M',  E'  G1  and 
P'  M2,  as  shown  by  K'  O'  F1  O2,  will  give  the  shape  of 
iiiik'r  line  K  K  in  plan. 


As  J  G  II  I  of  elevation  is  without  flare  the  pat- 
tern for  this  part  is  procured   in   the   ordinary  manner 
as  for  a  pieced  elbow.     Since  J'  M'  G'  M"  is  the  profile 
of  an  elliptical  cylinder,  the  section  upon  the  oblique 
line  J  G  of  the  elevation,  cut- 
ting the  same,  must  necessarily 
be  an  ellipse  whose  major  axis 
is  equal  to  M'    M3  and  whose 
minor  axis  is  equal  to  J  G.     In 
like  manner,  the  section  at  K  F 
of  the  elevation  may  be  assumed 
as  an  ellipse  whose  minor  axis 
is  K  F  and  whose  major  axis  is 
equal  to  O'  0'. 

In  Fig.  710,  a  duplicate  of 
K  F  G  J  of  elevation  is  shown 
by  T  R  U  W.  Bisect  T  E  in 
d  and  erect  the  perpendicular  d 
S,  and  make  d  S  equal  to  Q'  0' 
of  plan.  Through  the  points  T, 
S  and  R  trace  the  half  ellipse, 
as  shown.  In  a  similar  manner 
bisect  W  U  in  I,  and  erect  the 
perpendicular  I  V,  in  length 
equal  to  N1  M1  of  plan.  Through 
the  points  thus  obtained  trace 
the  half  ellipse  W  V  U.  Divide 
T  S  R  into  any  convenient  num- 
ber of  equal  parts,  and  from  the 
points  thus  obtained  drop  per- 
pendiculars cutting  T  R,  as 
shown.  Also  divide  "W"  V  U 
into  the  same  number  of  equal 
parts  as  was  T  S  R,  and  from 
the  points  thus  obtained  drop 
perpendiculars  cutting  W  U. 
Connect  points  in  T  R  with 

those  opposite  in  W  U,  as  shown  by  the  solid  lines. 
Thus  connect  a  with  A,  b  withy,  c  with  &,  etc.  Also 
connect  the  points  in  T  R  with  those  in  W  0,  as  indi- 
cated by  the  dotted  lines.  Thus  connect  1  with  h,  a 
with  y,  b  with  &,  etc. 

The  next  step  will  be  to  construct  a  series  of  sec- 
tions upon  the  several  solid  and  dotted  lines  just  drawn 
for  the  purpose  of  obtaining  the  true  distances  which 
they  represent.  The  sections  represented  by  solid  lines 


404 


Tlie  New  Metal    Worker  Pattern  Book. 


are  shown  in  Fig.  711.     To  construct  these  sections 
proceed  as  follows  :     For  section  a,  draw  the  line  a  A, 


13 


Fig.  710. — Elevation  of  Middle  Section  of  Elbow,  Showing  Method 
of  Triangulation. 


21 


II1 


171 


12' 


161 


15' 


d 
Fig.  711.— Sections  on  Solid  Lines  of  Fig.  710. 

in  length  equal  to  a  A  of  Fig.  710.     From  a  erect  a 


by  a  2',  and  from  h  erect  a  perpendicular,  in  length  equal 
to  h  11  in  W  V  U,  as  shown  by  h  II1,  and  connect  2' 
11',  as  shown.  For  section  b,  drawft/,  in  length  equal 
to  bj  of  Fig.  710.  From  b  erect  the  perpendicular  b  3", 
in  length  equal  to  b  3  in  T  R  S,  and  from/  erect  a  per- 
pendicular, in  length  equal  to  j  12  in  W  V  U,  and  con- 
nect 31  12',  etc.  For  the  sections  representing  the 
dotted  lines,  as  shown  in  Fig.  '712,  proceed  in  a  similar 
manner.  For  section  A,  draw  1  A,  in  length  equal  to 
1  h  of  Fig.  710.  From  h  erect  a  perpendicular,  in  length 


h 

12" 


g     is 


132 


5" 


153 


0  i          d 

Fig.  712.— Sections  on  Dotted  Lines  of  Fig.  710. 

equal  to  A  11  in  W  V  U,  and  connect  1  with  11'.  For 
section  j,  draw  aj,  in  length  equal  to  a  j  of  Fig.  710. 
From  a  erect  the  perpendicular  a  23,  in  length  equal  to 
a  2  in  T  S  R,  and  from  /  erect  a  perpendicular,  equal  in 
length  toy  12  in  W  V  U,  and  connect  22  12',  etc. 

To  describe  the  pattern  for  part  of  article  repre- 
sented in  Fig.  710  by  T  R  U  W,  as  shown  in  Fig.  71:;, 
proceed  as  follows  :  Draw  any  line,  as  R  V  of  pattern, 


perpendicular,  in  length  equal  to  a  2  in  T  S  R,  as  shown  ;  in  length  equal  to  R  U  of  Fig.  710.     With  R  of  pat- 


I'ntti-rn  Problems. 


405 


tern  as  center,  and  1  II2  of  the  diagram  of  sections,  Fig. 
712,  as  radius,  describe  a  small  are,  which  intersect  with 
one  struck  from  point  V  of  pattern  as  center,  and  1"  1 1  of 
Fig.  Tin  as  radius,  thus  establishing  the  point  11  of 
pattern.  With  point  11  of  pattern  as  center,  and  2'  11' 
of  diagram  of  sections,  Fig.  711,  as  radius,  describe  a 


tances  between  points  represented  by  numbers  in  Fig. 
711  for  the  length  of  solid  lines  in  pattern,  and  the 
spaces  in  E  S  T  for  the  stretchout  of  R  T"  of  pattern. 
Lines  drawn  through  the  points  thus  obtained,  as  indi- 
cated  by  J{  T'  \V  V,  will  be  one-half  of  the  required 
pattern.  The  other  half,  as  shown  by  R  T  W  V,  can 


Fig.  713.— Pattern  of  Middle  Section  of  Elbow. 


small  arc,  which  intersect  with  one  struck  from  R  of 
pattern  as  center,  and  R  2  of  Fig.  710  as  radius,  thus 
establishing  point  2  of  pattern.  Proceed  in  this  man- 
ner, using  the  dotted  lines  in  Fig.  712  for  the  distances 
in  pattern  represented  by  dotted  lines;  the  spaces  in 
\\  V  U  for  the  stretchout  of  V  W  of  pattern;  the  dis- 


be  obtained  by  a  repetition  of  the  same  process  or  by 
duplication. 

The  pattern  for  E  F  K  L  of  elevation  can  be  ob- 
tained in  a  similar  manner,  A  B  C  of  profile  being  one- 
half  the  shape  on  E  L,  and  R  S  T  of  Fig.  710  being 
the  half  section  on  F  K. 


PROBLEM    213. 


Patterns  for  a  Right  Angle  Piece  Elbow  to  Connect  a  Round  with  a  Rectangular  Pipe. 


In  Fig.  714  is  shown  the  design  of  a  right  angle 
elbow  of  which  one  end  is  rectangular,  as  shown  by 
N  O  P  Q,  and  the  other  round,  as  shown  by  A  B  C  D. 
Such  an  elbow  may  be  constructed  in  any  number  of 
pieces,  the  elevation  for  which  may  be  drawn  in  the 
manner  described  in  the  case  of  an  ordinary  piece 
elbow. 

In  the  present  instance  the  elbow  consists  of  seven 
pieces.  In  adjusting  the  transition  from  the  rectangle 
to  the  circle,  it  is  evident  that  the  flat  sides  of  each  of 
the  five  intermediate  pieces  must  become  shorter  in 
each  piece  as  the  round  end  of  the  elbow  is  approached ; 
and  that  the  quarter  circles  forming  the  corners  of  each 
of  the  intermediate  pieces  must  be  of  shorter  radius  as 
the  corners  of  the  rectangular  end  are  approached. 


This  may  be  accomplished  upon  the  elevation  in  the 
following  manner :  Through  the  center  of  the  eleva- 
tion draw  the  lines  c  to  m,  and  divide  L  T  into  the 
number  of  parts  there  are  pieces  in  the  elbow  subjected 
to  the  change  in  shape,  in  the  present  instance  five. 
From  k'  set  off  each  way  four  spaces,  as  shown  by  k  k' 
and  k'  k".  Set  off  from  f  three  spaces,  as  shown  by 
/ /'  and//'.  Continue  this  operation  and  connect  the 
points  L  k  j  h  y  f  and  g"  h"  f  k"  I/,  thus  showing  in 
side  elevation  the  change  from  the  rectangle  N  O  P  Q 
to  the  circle  A  B  C  D. 

To  show  a  similar  shape  on  the  outer  curve  of 
elbow,  draw  any  line,  as  E  M,  in  Fig.  715.  From  E 
on  E  M  set  off  the  spaces  E  F,  F  G,  etc.,  to  L  M  of 
Fig.  714.  As  it  is  only  necessary  to  show  the  half 


406 


Tlte  New  Metal   Worker  Pattern  Boole. 


shapes,  from  M  and  L  erect  perpendiculars,  in  length 
equal  to  N  n  of  profile,  and  connect  same,  as  shown 
by  M''  L''.  Connect  L2  F,  and  from  the  points  in  E  M  I 


will   indicate  the  method  to  be 

.sections. 


followed  in  the  other 


ELEVATION 
K 


PROFILE 


rf.  714.— Elevation  and  Profiles  of  an  Elbow  to  Connect  a  Hound 
With,  a  Rectangular  Pipe. 


erect  perpendiculars  cutting  F  L*.  The  shapes  on  in- 
ner curve  of  elbow,  as  shown  in  Fig.  716,  are  obtained 
in  the  same  manner  as  described  for  Fig.  715. 

For  the  hights  of  section  on  c  m  of  the  elevation, 
on  any  line,  as  E  M,  in  Fig.  717,  starting  from  E,  set 
off  the  distances  cf,fy',  g'  h',  etc.,  and  from  the  points 
thus  obtained  erect  perpendiculars,  as  shown.  From 
E  and  F  set  off  the  distance  C'  B  of  profile  of  circular 
end  and  draw  E'  F"  From  M  and  L  set  off  the  dis- 
tance N  n  of  profile  of  rectangular  end  and  draw  Ms 
L*.  Connect  L3  with  F3,  thus  completing  the  section. 

The  method  for  obtaining  the  patterns  for  sections 
E  F  F'  E'  and  L  M  M'  L'  is  the  same  as  for  an  ordinary 
pieced  elbow.  The  method  for  obtaining  the  pattern 
for  one  of  the  remaining  sections  will  be  shown,  which 


In  Fig.  718,  F  G  G'  F'  is  a  duplicate  of  the  sec- 
tion having  similar  letters  in  elevation. 
0  The  shape  F    II    F'    is    the    half    of  an 

ellipse,  because  F'  F  of  Fig.  71-t  is  an 
oblique  section  of  a  cylinder  of  .which 
A  B  C  D  is  the  plan,  and  can  be  de- 
scribed in  any  convenient  manner. 

From  G  erect  the  perpendicular  G 
S,  equal  to  G  G'  of  Fig.  715,  and  from 
G'  erect  another  perpendicular,  equal  to 
G  G'  of  Fig.  716.  From  points  g  and  y' 
erect  perpendiculars  equal  to  G3  G  of 
Fig]  717,  and  connect  T  T',  as  shown. 
p  Connect  S  T  and  T'  S'  by  a  quarter  of 

an  ellipse.  Divide  S  T,  T'  S',  F  U  and 
U  F'  into  any  convenient  number  of  equal  parts,  and 
from  the  points  thus  obtained  drop  perpendiculars  cut- 


G 


Fig.  715.— Shape  of  Flat  Part  of     Fig.  716.— Shape  of  Flat  Part  of 
Outer  Curve  of  Elbow.  Inner  Curve  of  Elbow. 


ting  G  G'  and  F  F'.     Connect  these  points,   as  shown 
by  the  solid  and  dotted  lines  in  G  G'  F'  F. 

The  next  operation  will  consist  in  constructing 
sections  upon  these  solid  and  dotted  lines  for  the  pur- 


Pattern   Problems. 


407 


pose  of  ascertaining  the  correct  distances  which  they 
represent.  These  sections  are  shown  in  Figs.  Tl'J  and 
720.  To  construct  the  sections  represented  by  solid 


Fig.  718.— Elevation  of  One  of  the  Pieces,  Showing  Method  of 
Trianyulation. 


lines  in  Fig.  718,  proceed  as  follows:  Draw  the  line 
F  G  of  Fig.  719,  in  length  equal  to  F  G  of  Fig.  718, 
and  from  point  G  erect  a  perpendicular,  in  lengtlrequal 
to  G  S  of  Fig.  718.  Connect  F  S,  which  gives  the 
distance  between  points  F  and  S.  For  the  second 
section  draw  v  o,  in  length  equal  to  v  o  of  Fig.  718, 


H3f 


Fig.  717.— Extended  Section  on  Line  c  m  of  Fig.  714. 

and  make  the  perpendiculars  o  6  and  v  7  of  the  section 
equal  to  lines  having  similar  letters  in  Fig.  718.  The 
other  sections  are  obtained  in  a  similar  manner.  To 


construct  the  sections  represented  by  dotted  lines  in 
Fig.  718,  proceed  as  follows:  Draw  the  line  F  o  of 
Fig.  7i'i,  ia  length  equal  to  F  o  of  Fig.  718,  and  from 
o  erect  the  perpendicular  o  6,  equal  to  o  (5  of  Fig.  718, 
and  connect  F  6.  For  the  second  section  draw  v p,  in 
length  equal  to  vp  of  Fig.  718,  and  from  points  v  and 
p  erect  perpendiculars  equal  to  v  7  and  p  5  of  Fig.  718. 
Connect  points  7  5,  which  give  the  distance  between 
corresponding  points  in  Fig.  718. 

In  Fig.  721,  G  G'  F'  F  is  the  pattern  of  part  of 


71    G 


V 

F" 


Fiii.  719.— Sections  on  Solid  Lines  of  Fig.  718. 

article  shown  by  similar  letters  in  Figs.  714  or  718. 
The  distances  represented  by  solid  lines  in  pattern  are 
obtained  from  the  sections  in  Fig.  719,  as  indicated  by 
corresponding  figures,  and  the  distances  represented  by 
dotted  lines  in  pattern  are  obtained  from  the  sections 
in  Fig.  720.  The  stretchout  of  G  G'  of  the  pattern  is 
obtained  from  G  T  T'  G' of  Fig.  718,  as  the  stretchout 
of  F  U  F'  of  the  pattern  is  obtained  from  F  U  F'  of 
Fig.  718. 


408 


The  New  Metal   Wwker  Pattern  Book. 


To  develop  the  pattern  from  tlie  sections  above 
constructed,  first  draw  G  F  of'Fig.  721,  in  length  equal 
to  G  F  of  Fig.  718  or  719,  upon  which  duplicate  the 


101- 


p 


Fig.  720. — Sections  on  Dotted  Lines  of  Fig.  718. 

triangle  G  F  S  of  Fig.  719,  as  shown.  From  S  of  pat- 
tern as  center,  with  S  6  of  Fig.  718  as  radius,  de- 
scribe an  arc,  6,  which  intersect  with  one  struck  from 


F  of  pattern  as  center,  with  radius  F  6  of  Fig.  720, 
thus  establishing  point  6  of  pattern.  Then  with  radius 
F  7  of  Fig.  718,  from  F  of  pattern  as  center,  describe 
an  arc,  which  intersect  with  a  second  arc  struck  from 
point  6  of  pattern  as  center,  and  0  7  of  Fig.  719  as 
radius,  thus  establishing  point  7  of  pattern.  Continue 
this  process  until  the  various  points  indicated  in  pat- 
tern are  located.  Lines  drawn  through  the  points 


S'G' 


F  7 

Fig.  721.— Half  Pattern  of  F  G  G'  F'  of  Fig.  714. 

thus  obtained,  as  indicated  by  G  S  T  T'  S'  G'  and  F' 
U  F,  will  complete  one-half  of  the  required  pattern. 
The  other  half  can  be  obtained  by  duplication  or  by  a 
repetition  of  the  above  process. 

In  obtaining  the  pattern  for  anv  one  of  the  re- 
maining pieces  first  draw  a  duplicate  of  its  elevation  as 
taken 'from  Fig.  714,  upon  either  side  of  which  con- 
struct the  proper  section,  obtaining  the  points  in  the 
same  from  Figs.  715,  71  <>  and  717  as  was  done  in  Fig. 
718,  after  which  the  subsequent  operation  is  analogous 
to  that  above  described. 


PROBLEM  214. 

Pattern  for  the  Soffit  of  a  Semicircular  Arch  in  a  Circular  Wall,  the  Soffit  Being  Level  at  the  Top  and 
the  Jambs  of  the  Opening  Being  at  Right  Angles  to  the  Walls  in  Plan.    Two  Cases. 


First  Case. — In  Fig.  722,  let  A  B  C  represent  the 
outer  curve  of  an  arch  in  a  circular  wall  corresponding 
to  A'  H  C'  of  plan,  and  let  E  B  D  represent  the  inner 
opening  in  the  wall,  as  shown  by  E'  F'  D'  in  plan. 
Then  A  E  B  C  D  will  represent  the  soffit  of  the  arch 
in  elevation  and  A'  H  C'  D'  F'  E'  the  same  in  plan. 
In  the  engraving  the  outer  curve  of  the  arch  is  a  per- 
fect semicircle,  and  the  inner  curve  is  stilted,  as 
shown,  so  as  to  make  the  soffit  level  at  B.  Instead 
of  the  stilted  arch,  the  inner  curve  may,  if  desired,  be 
drawn  as  a  semi- ellipse  of  which  E  D  is  the  minor 
axis  and  F  B  one-half  of  the  major  axis. 

Divide  A  B  of  elevation  into  any  convenient  num- 
ber of  equal  parts,  shown  by  the  small  figures.  With 


the  T-square  parallel  with  the  center  line  B  B',  drop 
lines  from  the  points  in  A  B,  cutting  A'  II  of  plan. 
as  shown.  Since  that  portion  of  the  inner  arch  from 
E  to  12  is  drawn  vortical,  as  above  explained,  divide 
12  B  into  the  same  number  of  parts  as  was  A  B,  and, 
with  the  T-square  parallel  with  the  center  line  B  B', 
drop  lines  to  E'  F',  as  shown.  Connect  opposite 
points  in  A'  II  with  those  in  K  !•".  as  shown  l>v  the 
solid  lines  in  plan.  Also  divide  the  four-sided  figures 
thus  produced  by  means  of  the  diagonal  dotted  lines 
6  8,  5  9,  etc.,  as  shown.  The  several  triangles  thus 
produced  will  represent  in  plan  the  triangles  into  which 
the  soffit,  or  under  side,  of  the  arch  is  divided  for  the 
purpose  of  obtaining  its  pattern.  In  order  to  ascer- 


1'attern  Problems. 


409 


tain  the  real  Jistances  across  tlie  surface  of  the  arch 
which  the  solid  and  dotted  lines  represent,  it  will  be 
necessary  to  construct  a  series  of  sections  of  which 
these  lines  are  the  bases,  as  shown  in  Figs.  723  and 
734. 

In  constructing  the  diagram  shown  Fig.  723,  the 
several  solid  lines  of  the  plan,  though  not  exactly 
equal  in  length  (because  tliev  an!  not,  drawn  radiallv 
from  the  center  of  the  curve  A'  II  C'),  may  be  con- 
sidered as  of  the  same  length.  Draw  the  right  angle 
P  Q  R  as  in  Fig.  723,  and  from  Q  set  off  horizontally 
the  distance  II  F'  of  plan,  as  shown  by  Q  R.  Draw 
R  S  parallel  with  Q  P,  and,  measuring  from  Q,  set  off 
on  Q  P  the  length  of  lines  dropped  from  points  in  A 

ELEVATION 
B 


fig.  722. — Plan  and  Elevation  of  Arch  in    a    Circular    Wall.— 
first  Case. 

B  to  A  F,  as  shown  by  corresponding  figures  2  to  6. 
Likewise  set  off  from  R  on  R  S  the  length  of  lines 
dropped  from  points  in  E  B  to  E  F,  as  shown  by  the 
figures  12  to  7,  and  connect  the  points  in  P  Q  with 
those  in  S  R,  as  indicated  by  the  solid  lines  in  plan. 
Thus  connect  1  with  12,  2  with  11,  3  with  10,  etc. 
To  construct  the  diagram  based  upon  the  dotted  lines 
of  the  plan,  draw  the  right  angle  M  N  0  in  Fig.  724, 
and,  measuring  in  each  instance  from  N,  set  off  on  N 
M  the  same  distances  as  in  Q  P  of  Fig.  723.  Starting 
from  N,  set  off  on  N  0  the  lengths  of  dotted  lines  in 
plan,  as  shown  by  the  small  figures  in  N  0.  With 
the  T-square  parallel  with  M  N,  draw  lines  from  the 


points  in  N  0,  and,  in  each  instance  measuring  from 
N  0,  make  these  lines  of  the  same  length  as  lines  of 
similar  number  dropped  from  points  in  E  B  of  eleva- 
tion to  E  F.  Connect  the  points  in  these  lines  with 
points  in  M  N,  as  indicated  in  plan  by  the  dotted 


I] 
10 


1211-," 

to 


Fig.  723.— Diagram  of  Sections  an     Fig.  724.— Diagram  of  Sections  on 
Solid  Lines  of  Plan,  fig.  722.  Dotted  Lines  of  Plan,  fig.  722. 


lines.  Thus  connect  6  with  8,  5  with  9,  4  with 
10,  etc. 

The  next  step  is  to  obtain  the  distances  between 
points  in  A  B  of  elevation  as  if  measured  on  the  outer 
opening  in  the  curved  wall.  To  do  this,  on  F  A  ex- 
tended set  off  a  stretchout  of  A'  H  of  plan,  as  shown 
by  the  small  figures  5',  4',  etc.,  and  with  the  T-square 
at  right  angles  to  the  stretchout  line  J  F,  draw  the 
usual  measuring  lines.  With  the  "[-square  parallel 
with  J  F,  carry  lines  from  the  points  in  A  B  to  lines 
of  similar  number  drawn  from  the  stretchout  line.  A 
line  traced  through  these  points,  as  shown  by  J  B,  will 
give  the  true  distances  desired  between  the  points  in 
the  outer  curve  of  the  arch. 

The  distances  between  points  in  E  B,  the  inner 
curve,  are  obtained  in  a  similar  manner.  To  avoid  a 
confusion  of  lines,  the  stretchout  of  F'  E'  of  plan  is 


13  12  11 

Fig.  125.— One-Half  Pattern  of  Soffit  of  Arch  Shown  in  fig.  7%2. 

set  off  on  F  C  of  elevation,  as  shown  by  the  small 
figures,  7',  8',  9',  etc.  B  D  is  also  divided  into  the 
same  parts  as  was  E  B,  and  from  the  points  thus  ob- 
tained lines  are  drawn  to  the  right  parallel  with  F  C. 
With  the  T-square  parallel  with  B  F,  carry  lines 


410 


J7ie  New  Metal   Worker  Pattern  Book, 


from  the  stretchout  points  in  F  K,  cutting  lines  of 
similar  number  drawn  from  the  points  in  B  D.  A 
line  traced  through  the  points  thus  obtained,  as  shown 
by  B  K,  will  give  the  distance  between  points  as  if 
measured  on  the  inner  curved  line  of  the  wall. 

From  the  several  sections  now  obtained  the  pat- 
tern may  be  developed  in  the  following  manner :  At 
any  convenient  place  draw  the  line  a  e  in  Fig.  725, 
making  it  in  length  equal  to  Q  K  of  Fig.  723,  or  A'  E' 


thus  establishing  the  point  11  of  pattern.  Continue 
in  this  way,  using  the  tops  of  the  sections  in  Figs. 
7  •_'.">  ami  724  for  measurements  across  the  pattern,  the 
spaces  in  J  B  for  the  distances  along  the  edge  a  h  of 
pattern,  and  the  spaces  in  B  K  for  the  distances  along 
the  inner  edge  e  f,  establishing  the  several  points,  as 
shown.  Through  the  points  in  a  h  and  e  f  lines  are 
to  be  traced,  while  f  h  is  to  be  connected  by  a  straight 
line,  thus  completing  one-half  the  pattern.  The  other 


123  4  5  6|< 

DEVELOPMENT  OF 
OUTER  CURVE 


PLAN 
Fig.  726.— Plan  and  Elevation  of  Arch  in  a  Circular  Wall.— Second  Case. 


of  plan.  At  right  angles  to  a  e  draw  e  12,  in  length 
equal  to  R  12  of  Fig.  723,  and  connect  a  with  12  if  it 
is  desired  to  show  the  triangle.  From  a  as  center,  and 
J  2  of  elevation  as  radius,  describe  a  small  arc,  2, 
which  intersect  with  one  struck  from  point  12  of  pat- 
tern as  center,  and  12  2  of  Fig.  724  as  radius,  thus  es- 
tablishing the  point  2  of  pattern.  With  2  of  pattern 
as  center,  and  2  11  of  Fig.  723  as  radius,  describe 
another  arc,  11,  which  intersect  with  one  struck  from 
12  of  pattern  as  center,  and  12  11  of  B  K  as  radius, 


half  of  pattern  can  be  obtained  by  the  same  method 
or  by  any  convenient  means  of  duplication. 

If  the  arch  were  semi-elliptical  instead  of  semi- 
circular, the  method  of  procedure  would  be  the  same 
as  above  described. 

Second  Case. — In  Fig.  726,  ABC  represents  the 
outer  curve  of  an  arch  in  circular  wall,  as  shown  by 
A'  H  C'  in  plan.  E  B  D  represents  the  inner  curve 
in  elevation,  as  does  E'  D'  the  same  in  plan.  Thru 
A  E  B  D  C  represents  the  soffit  of  the  arch  in  eleva- 


Pattern   Proilt  in*. 


411 


tion  and  A'  II  0'  D'  E'  the  same  in  plan.  The  con- 
ditions given  in  this  ease  differ  from  those  of  the  first 
case  only  in  the  fact  that  the  inner  curve  of  the  arch 
in  this  ease  is  straight  in  plan,  as  shown  by  E'  D', 
instead  of  curved  to  the  radius  of  the  wall  as  in  Fig. 
722.  The  method  of  procedure  in  this  case  is  exactly 
the  same  as  before,  but  one  less  operation  will  lie.  nec- 
essary, since  measurements  upon  the  inner  curve  may 
lie  taken  directly  from  E  B  D  of  the  elevation. 

To    avoid    a    confusion    of    lines,    a    duplicate 


ef 

4' 
3' 

2' 

"  1  2     345" 




—  1* 

"""— 

3 

---,2' 

l| 

12  3    456 


fiij.  7-'7.—I>i(iyraui  of  Sections  on       Fig.  728. — Din  gram  of  Sections  on 
Solid  Lines  of  Plan,  Fiij.  796.  Dotted  Lines  of  Plan,  Fig.  7K. 

of  E  B  I)  of  elevation  has  been  drawn  in  plan,  as 
shown  bv  E'  B'  D'.  To  obtain  the  divisions  on  plan 
divide  A  B  into  any  convenient  number  of  equal  parts, 
and  from  the  points  thus  obtained  drop  lines  parallel 
with  the  center  line  B  B'  to  A'  II  of  plan,  as  shown. 
Divide  E'  B'  in  a  similar  manner,  and  from  the  points 
thus  obtained  drop  lines  to  E'  F'  of  the  plan,  as  shown. 
Connect  points  in  A'  II'  with  those  of  similar  number 
in  E'  F'  by  solid  lines.  Also  connect  points  in  A'  H 
with  those  of  next  lower  number  in  E'  F'  by  dotted 
lines.  These  solid  and  dotted  lines  just  drawn  will 
.form  the  bases  of  a  series  of  sections,  shown  in  Figs. 
727  and  728,  whose  upper  lines  will  give  correct  dis- 
tances across  the  pattern  of  the  soffit. 

To  construct  the  sections  based  upon  the  solid 
lines  of  the  plan,  first  draw  the  right  angle  P  Q  R  in 
Fig.  727,  and  set  off  on  Q  P,  measuring  from  Q,  the 
length  of  the  vertical  lines  in  A  B  F  of  elevation. 
Starting  from  Q,  set  off  on  Q  R  the  length  of  solid 
lines  in  A'  H  F'  E'  of  plan,  as  shown  by  the  small 
figures  in  Q  R.  With  the  T-square  parallel  with  P  Q, 
draw  lilies  from  the  points  in  Q  R,  and,  measuring 
from  Q  R,  set  off  on  these  lines  the  length  of  lines  of 
corresponding  number  in  E'  F'  B'  of  plan,  and  connect 
the  points  with  points  of  similar  number  in  P  Q.  The 
diagram  of  sections  based  upon  the  dotted  lines  of  the 
plan,  shown  in  Fig.  728,  is  constructed  in  the  same 


manner,  using  the  length  of  dotted  lines  in  plan  for 
the  distances  in  N  0,  the  length  of  lines  in  E'  F'  B'  of 
plan  for  the  length  of  lines  set  off  at  right  angles  to 
N  0,  and  the  length  of  lines  in  A  B  F  of  elevation 
for  the  distances  in  M  N.  Connect  the  points  as  in- 
dicated by  the  dotted  lines  of  the  plan,  all  as  shown. 

The  next  step  is  to  obtain  the  correct  distances 
between  points  in  A  B  of  elevation,  or  A'  II  of  plan. 
To  do  this  lay  off  horizontally  J  K,  on  which  set  off 
a  stretchout  of  A'  H  of  plan,  and,  with  the  T-square 
at  right  .ingles  with  J  K,  draw  the  usual  measuring 
lines.  With  the  T-square  parallel  with  J  K,  carry 
lines  from  the  points  in  A  B  to  lines  of  similar  num- 
ber. A  line  can  be  traced  through  these  points,  as 
shown  by  J  L,  from  which  the  correct  stretchout  of 
the  outer  side  of  the  pattern  can  be  obtained. 

To  describe  the  pattern  first  draw  any  line,  as  a  e 
of  Fig.  729,  equal  to  A'  E'  of  plan.  With  e  of  pattern 
as  center,  and  E'  1  of  the  inner  curve  of  the  arch  as 
radius,  strike  a  small  arc,  1',  which  intersect  with  one 
struck  from  a  of  pattern  as  center,  and  Q  1'  of  Fig. 
7'27  as  radius,  thus  establishing  the  point  1'  of  pat- 
tern. With  a  of  pattern  as  center,  and  J  2  of  Fig. 
72<i  as  radius,  describe  a  small  arc,  2,  which  intersect 
with  one  struck  from  1'  of  pattern  as  center,  and  1'  2' 
of  Fig.  728  as  radius,  thus  establishing  point  2  of 
pattern.  Then  from  2  as  center,  with  2'  2'  of  Fig. 
727  as  radius,  strike  a  small  arc,  2',  which  is  inter- 


C  5' 

Fin.  Ti'J.—Half  Pattern  of  Soffit  Shown  in  Fig.  726. 

sected  with  one  struck  from  1'  of  pattern  as  center, 
and  1  2  of  the  inner  curve  of  the  arch  as  radius,  thus 
definitely  establishing  the  point  2'  of  pattern.  Con- 
tinue in  this  way,  using  the  tops  of  sections  in  Figs. 
727  and  728  for  measurements  across  the  pattern,  and 
the  spaces  in  the  inner  curve  E'  B'  and  in  outer  curve 
as  developed  in  J  L.  Fig.  726,  for  the  distances  about 
the  edges  of  the  pattern,  establishing  the  several 
points,  as  shown.  Connect  points  h  and  /  with  a 
straight  line,  and  through  the  intermediate  points  trace 
the  curves,  as  shown.  The  other  half  of  pattern  can 
be  obtained  by  the  same  method  or  by  duplication. 


412 


The  New  Metal    Worker  Pattern  Hook. 


PROBLEM  215. 

Pattern  for  a  Splayed  Elliptical  Arch  in  a  Circular  Wall,  the  Opening  Being  Larger  on  the  Outside 

of  the  Wall  than  on  the  Inside. 


In  Fig.  730  is  shown  flu1  elevation  and  plan  of 
an  elliptical  window  head  in  a  circular  wall.  The 
outer  curve  of  head  is  represented  bv  ABC  in  elevation 
and  by  A'  B'  C'  in  plan.  The  inner  curve  is  repre- 
sented by  F  E  D  in  elevation  and  F'  E'  D'  in  plan. 
A  B  C  D  E  F  therefore  represents  the  splaved  or 
beveled  portion  in  elevation  for  which  the  pattern  is 
required,  and  A'  B'  C'  D'  E'  F'  the  plan  of  the  same. 
Divide  A  B  of  elevation  into  any  convenient  number 


of  sections  <>n  A'  J5  of  plan,  as  the  lines  dropped  from 
F  E  to  F  G  give  the  hight  of  sections  on  F'  E'  of 
plan. 

To  construct  the  sections  shown  in  Pnig.  7-!l,  rep- 
resented in  plan  by  the  solid  lines,  proceed  as  follows: 
Draw  the  right  angle  a  a  1,  making  «  a'  equal  to  E'  B' 
of  plan,  and  a'  1  equal  to  G  B  (n  1)  of  elevation.  Dra\\ 
a  1  •-'  parallel  to  <i'  1,  making  its  length  equal  to  (i  K 
(a.  1  •_' i  of  elevation,  and  connect  12  with  1.  The  dis- 


H 


FIIJ.  7M. — Elevation  and  Plan  of  tiplayed  Elliptical  Arch. 


of  parts,  in  the  present  instance  five.  For  convenience 
number  the  points  thus  obtained;  as  shown  by  the 
small  figures.  Drop  perpendiculars  from  the  points 
irr  A  B,  cutting  A'  B'  of  plan,  as  shown.  Also  divide 
F  E  of  elevation  into  the  same  number  of  parts  as  was 
A  B,  and  drop  similar  perpendiculars  to  F'  E'  of  plan. 
Connect  opposite  points  in  A'  B'  with  those  in  F'  E', 
as  shown  by  the  solid  lines,  as  2  11,  3  10,  etc.  Also 
divide  the  four-sided  figures  thus  produced  into  tri- 
angles by  means  of  the  dotted  lines,  as  shown  from  1 
to  11,  2  to  10,  etc.  To  ascertain  the  true  distances 
across  the  face  of  the  arch  which  these  several  solid 
and  dotted  lines  of  the  plan  represent,  it  will  be  nec- 
essary to  construct  vertical  sections  through  the  arch 
upon  each  one  of  these  lines  as  a  base.  The  lines 
dropped  from  A  B  to  A  G  of  elevation  give  the  hight 


tance  12  1  of  this  section  represents  the  actual  distance 
between  the  points  1  and  12  in  the  elevation  or  plan. 
The  second  section  is  constructed  in  a  similar  manner; 
It  c  represents  the  distance  2  11  of  plan;  I  11  is  equal 
to  li  11  of  elevation  and  c  '2  to  c  •>  of  elevation.  Con- 
nect 11  with  -2  of  section,  which  will  give  the  -ictnal 
distance  between  points  11  and  2  of  elevation  or  plan. 
The  remaining  sections  are  constructed  in  a  similar 
manner,  each  of  the  sections  representing  a  vertical 
section  through  the  head  on  the  lines  of  corresponding 
numbers  in  the  plan.  The  sections  based  upon  the 
.lotted  lines  of  the  plan,  shown  in  Fig.  V32,  are  con- 
structed in  exactly  the  same  manner.  Draw  It  a,  in 
length  equal  to  1  1 1  of  plan,  and  erect  the  two  perpen- 
diculars, as  shown.  Make  a  1  of  section  equal  to  //  I 
of  elevation,  and  also  make  I  11  of  section  equal  to  I 


Pattern  Problems. 


413 


11  of  elevation,  and  connect- 11  with  1.      The  remain- 
ing sections  are  constructed  in  the  same  manner. 

Before  ilie  pattern  can  be  obtained  it  will  be  nec- 
essary k>  develop  extended  sections  of  the  inner  and 


b 

4 


e  j  u  k 

Fiij.  "•',!. — Sections  Baaed  I'poti  the  Solid  Lines  of  the  Plan,  Fig.  730. 

outer  curves,  as  shown  to  the  left  and  right  of  the  ele- 
vation. This  is  done  for  the  purpose  of  obtaining  the 
actual  distance  between  points  shown  in  elevation. 
For  convenience,  on  (I  A  extended,  as  H  K,  lay  off  a 
stretchout,  of  A'  B'  of  plan,  and  from  the  points  therein 
contained  erect  the  perpendiculars,  as  shown.  From 
the  points  in  A  B  of  elevation  draw  lines  parallel  with 
H  G,  cutting  perpendiculars  of  similar  number  erected 
from  H  K.  A  line  drawn  through  these  points  of  in- 


a  d 

3 


e  f  k 

Fig.  732.—  Sections  Based  Upon  the  Dotted  Lines  of  the  Plan,  Fig.  730. 

tersection,  as  shown  by  II  J,  will  show  the  shape  of 
A  B  of  elevation  as  laid  out  on  a  flat  surface.  The 
development  of  the  inner  curve  is  shown  to  the  right 


of  the  elevation.  On  L  N  is  laid  out  the  stretchout  of 
F'  E'  of  plan,  and  on  the  perpendiculars  erected  from 
the  points  in  the  line  are  set  off  the  same  distances  as  on 
lines  of  similar  number  in  F  E  G  of  elevation.  A  line 
traced  through  these  points,  as  shown  by  L  M,  also 
shows  the  shape  of  F  E,  as  laid  out  on  a  flat  surface, 
and  gives  the  distance  between  points  as  if  measured 
on  the  finished  article. 

To  obtain  the  pattern,  using  the  distances  between 
points  in  II  J  and  L  M  of  Fig.  730,  and  the  diagrams 
in  Figs.  731  and  732,  proceed  as  follows:  Draw  any 
line,  as  B  E  in  Fig.  733,  in  length  equal  to  1  12  of  the 
first  section  in  Fig.  731.  With  E  as  center,  and  Mil 
of  inner  curve  as  radius,  describe  a  small  arc,  11,  which 
intersect  with  one  struck  from  B  as  center,  and  1  11  of 
Fig.  732  as  radius,  thus  establishing  the  point  11  of 
pattern.  With  11  of  pattern  as  center,  and  11  2  of 
the  second  section  in  Fig.  731  as  radius,  describe  an- 


Fig.  733.— Pattern  of  Splayed  Arch  in  Fig.  730. 

other  small  arc,  2,  which  intersect  with  one  struck 
from  point  B  of  pattern  as  center,  and  J  2  in  J  H  of 
outer  curve  as  radius,  thus  establishing  the  point  2  of 
pattern.  Continue  in  this  way,  using  the  tops  of  the 
sections  in  Figs.  731  and  732  for  the  measurements 
across  the  pattern,  the  spaces  in  the  inner  curve  L  M 
and  in  the  outer  curve  H  J  for  the  distances  about  the 
edges  of  the  pattern,  establishing  the  several  points, 
through  which  draw  the  lines  shown.  Then  B  A  F  E 
is  the  half  pattern  of  splayed  head,  shown  in  elevation 
by  A  B  E  F.  The  other  half  can  be  obtained  by  any 
convenient  means  of  duplication. 

A  semicircular  splayed  arch  can  be  developed  in 
the  same  manner  as  above  described. 

The  pattern  for  a  blank  for  a  curved  molding, 
either  semicircular  or  semi-elliptical,  for  an   arch  in  a ' 
circular  wall  comprises  really  the    same  relations  of 
parts  as  are  shown  in  Fig.  730,  and  could  be  obtained 
as  above  described. 


414 


77(e  New  Metal    Worker  Pattern  Book. 


PROBLEM   216. 

Pattern  for  a  Splayed  Arch  in  a  Circular  Wall,  the  Larger  Opening  Being  on  the  Inside  of  the  Wall. 


In  Fig.  734  is  shown  the  plan  and  elevation  of  an 
arch  in  a  curved  wall,  such  as  might  be  used  as  the 
head  of  a  window  or  door,  the  jambs  and  head  to  have 
the  same  splay.  In  the  plan,  C  E  D  represents  the 
inner  curve  of  wall  and  A  F  B  the  outer.  J  G  M  in 
elevation  represents  the  inner  curve  and  K  II  L  iho 
outer.  In  order  to  arrive  at  a  system  of  triangles  by 
means  of  which  to  measure  the  splayed  surface,  lirst 


the  inner  and  outer  lines  of  plan,  as  shown.  The  solid 
and  dotted  lines  of  the  plan  will  form  the  bases  of  a 
series  .if  right  angled  triangles  whose  altitudes  can 
be  found  in  the  elevation,  and  whose  h  vpothcmises 
will  give  correct  measurements  across  the  face  of  the 
splayed  surface. 

To  determine  the  bights  of  these  triangles  extend  to 
the  left  the  horizontal  lines  drawn  through  Iv  II  of  the 


ELEVATION 
G 


-H-l— -I 


<4  ~~  "L 


Ji  pflj  ill 

1 1        III 


DEVELOPMENT  OF 
INNER  CURVE 


DEVELOPMENT   OF 
OUTER   CURVE 


B 


PLAN 

Fig.  734. — Plan,  Elevation  and  Extended  Sections  of  Splayed  An-h. 


divide  J  G  of  the  elevation  into  any  convenient  num- 
ber of  parts,  in  this  case  five,  as  indicated  by  the  small 
figures.  From  the  points  in  the  curve  thus  established 
drop  lines  parallel  to  the  center  line  G  F,  cutting  the 
inner  curve  of  the  plan  C  E,  as  shown.  Next  carry 
lines  from  points  in  J  G  in  the  direction  of  the  center 
X,  intersecting  the  outer  curve  and  establishing  the 
points  7,  8,  9,  etc.,  in  it.  From  these  points  drop  lines 
to  the  outer  curve  in  plan  A  F,  establishing  the  points 
7,  8,  9,  etc.  Connect  opposite  points  in  J  G  with 
those  in  K  H,  as  1  and  7,  2  and  8,  3  and  9,  4  and  10, 
5  and  11,  and  6  and  12.  Likewise  connect  1  and  8, 
2  and  9,  3  and  10,  etc.,  as  shown  by  the  dotted  lines. 
In  the  same  manner  connect  corresponding  points  in 


elevation  until  they  intersect  lines  dropped  from  J  G, 
as  shown  by  the  points  b,  <1.  /and  //.  Then  //  -2.  <l  '•'. 
etc.,  will  be  the  required  bights.  To  construct  the  tri- 
angles which  represent  the  solid  lines  in  plan  and  ele- 
vation, proceed  as  shown  in  Fig.  7  .">.">:  For  the  lirst 
triangle,  G  II  7,  draw  the  right  angle  G  II  7,  making 
the  altitude  equal  to  G  II  of  elevation  and  the  base 
equal  to  E  F,  1  7,  of  plan,  and  draw  the  hypothennse. 
which  represents  the  actual  distance  between  the  points 
1  and  7  of  the  elevation  or  plan.  In  like  manner  the 
hypothcimso  of  the  second  triangle  258  shows  the 
actual  distance  between  the  points  2  and  S  of  the 
elevation;  '2  //  in  Fig.  73/5  is  made  equal  to  2  It  of  the 
elevation,  while  I  S  equals  2  8  of  the  plan. 


Pattern  Problems. 


415 


The  remaining  triangles  in  Fig.  735  arc  con- 
structed in  a  similar  manner,  each  of  the  triangles  rep- 
resenting a  Vertical  section  tlirongh  the  head  on  the 
lines  of  corresponding  numbers  in  the  plan. 

The  triangles  shown  in  Fig.  T.'iii  correspond  to 
similar  sections  taken  on  the  dotted  lines  in  plan.  The 
base  a  S  of  the  first  triangle  is  ei|iiul  to  1  S  of  the  plan, 
and  the  hight  1  a  to  1  a  of  elevation,  the  point  n  of 
the  elevation  being  on  a  level  with  s,  as  shown  bv 
the  dotted  line  s  a.  The  hypothennso  1  S  is  then 
drawn,  which  gives  the  distance  between  the  points  1 


Fig. 


"•In. — Diagram  of  Triangles  Rased  Upon  the  Solid  Lines  of 
the  Plan. 


and  8  in  plan  or  elevation.  The  bases  of  the  remain- 
ing triangles  are  derived  from  the  dotted  lines  in  plan, 
and  the  hights  from  the  distances  2  c,  3  e,  -i  y  and  5  £ 
of  elevation. 

Before   the    correct   measurements  or  stretchouts 
for  the  inner  and  outer  lines  of  the  pat- 
tern can  be  obtained  it  will  be  neces- 
sary to  develop  extended  sections  of  the 
inner  and  outer  curves  of  the  arch  in 
elevation,    as    shown    at    the    left  and 
right   of   that   view.      For  the  develop- 
ment of   the   outer  curve,  as  shown  to 
the   right  by  A'  H'  F'.  proceed  as  fol- 
lows :      On   X   M  extended    lay    off    a 
8tretch0ti1    equal   to  A- F  of  the  plan, 
transferring  it   point  by    point.      From 
the    points    thus    established  in    A'  F' 
carry  lines  vertically,   extending  them 
indefinitely,  as  shown,  and  then  from  the  points  in  the 
outer  curve  K  II  of  the  elevation  cany  lines  horizon- 
tally to  the  right,  intersecting  the  corresponding  lines 
just   drawn  from  A'  F',    and  through   the  points  thus  ' 
established    trace    the    curved    line  A'  II'.     The    de- 


velopment of  inner  curve,  as  shown  to  the  left,  is 
accomplished  in  a  similar  manner.  On  X  J  ex- 
tended lay  off  a  stretchout  of  C  E  of  plan,  and  from 
the  points  thus  established  carry  lines  vertically. 
From  the  points  in  the  inner  curve  J  G  carry  lines 
horizontally  to  the  left,  intersecting  lines  of  similar 
number,  and  through  the  points  thus  established  trace 
the  curve  C'  G'. 

To  describe  the  pattern  shown  in  Fig.  737,  first 


a 

3 


X 


\ 


Fig.  736. — Diagram  of  Triangles  Based  Upon  the  Dotted  Lines  of 
the  Plan. 

draw  the  line  g  h,  or  1  7,  in  length  equal  to  the  hypoth- 
enuse  1  7  of  the  first  triangle  in  Fig.  735.  From  7  as 
center,  with  7  8  of  the  development  of  the  outer  curve  as 
radius,  strike  a  small  arc,  as  shown  at  8  in  the  pattern. 
From  1  as  center,  with  1  8  of  the  first  triangle  in  Fig. 


PATTERN 
Fig.  737.— Pattern  for  Splayed  Arch. 


736  as  radius,  intersect  the  arc  last  struck,  thus  estab- 
lishing the  point  8.  From  1  as  center,  with  radius 
equal  to  1  2  of  the  development  of  inner  curve,  strike 
a  small  arc,  as  shown  a-t  2.  Then  from  8  as  center, 
with  8  2  of  the  second  triangle  in  Fig.  735  as  radius, 


410 


Tlie  New  Metal    Worker  Pattern  Book. 


intersect  the  arc  at  2  already  drawn,  thus  definitely 
establishing  the  point  2  in  the  upper  line  of  the  pattern. 
Continue  in  this  way,  using  the  hypothenuses  of  the 
several  triangles,  as  shown,  for  measurements  across  the 
pattern,  and  the  spaces  of  the  inner  and  outer  curves 


as  developed  in  Fig.  734  for  the  distances  along  the 
edges  of  the  pattern,  establishing  the  several  points,  as 
shown.  Lines  traced  through  the  points  from  c  to  <j 
and  from  «  to  h  will  complete  the  pattern  for  one-half 
the  arch.  The  complete  pattern  is  shown  in  Fig.  737. 


PROBLEM  217. 

Pattern  for  the  Soffit  of  an  Arch  in  a  Circular  Wall,  the  Soffit  Being  Level  at  the  Top  and  the 
Jambs  of  the  Opening  Being  Splayed  on  the  Inside. 


In  Fig.  738,  A  B  C  is  the  elevation  of  the  inner 
curve  A'  H  C'  of  the  plan  and  E  B  D  that  of  the  outer 
curve  E'  F'  D'.  As  will  be  seen  by  inspection,  the 
outer  curve  is  drawn  from  G  as  center,  that  portion  of 
the  opening  from  the  springing  line  down  to  thesprinu 


the  outer  curve  of  wall  is  struck  from  G  as  center, 
divide  12  B  into  the  same  number  of  parts  as  was  A  1>, 
and  drop  lines  from  these  points  to  E'  F'  of  plan,  as 
shown.  Connect  opposite  points  in  E'  F'  of  plan  with 
those  in  A'  II.  as  indicated  b\-  the  solid  lines.  Also 

ELEVATION 
B 


M  1211  10      9          8         ?N 

SHAPE  ON    E   F'oF  PLAN 


Fig.  7i8. — Plan,  Elevation  and  Extended  Sections  of  an  Arch  in   u 
Circular  Wall. 


ing  line  of  the  inner  curve,  as  E  12,  being  straight  and 
vertical.  The  pattern  of  the  soffit  could  have  been 
obtained  in  exactly  the  same  manner  had  the  elevation 
of  this  outer  curve  been  a  semi-ellipse. 

Divide  A  B  of  the  elevation  line  into  any  con- 
venient number  of  equal  parts,  and  with  the  "T-square 
parallel  with  center  line  B  H  drop  lines  from  the 
points  in  A  B,  cutting  A'  H  of  plan,  as  shown  by  the 
small  figures  1  to  6.  As  the  semicircle  representing 


connect  the  points  of  the  plan  obliquely,  as  shown  l>v 
the  dotted  lines,  thus  dividing  the  plan  of  the  soffit  of 
the  arch  into  triangles.  In  order  to  ascertain  the  true 
distances  which  these  lines  drawn  across  the  plan  repre- 
sent it  will  be  necessary  to  construct  a  series  of  sections^ 
of  which  they  are  the  bases,  as  shown  in  Figs.  I'.W  a.nd 
740. 

In  Fig.  739  is  shown  a  diagram  of  sections  corre- 
sponding to  the  solid  lines  in  plan,  to  construct  which 


Pattern  Problems. 


417 


proc-ivd  ;is  follows:  Draw  the  right  angle  P  Q  K, 
and,  measuring  from  0,  set  off  on  <i  P  the  length  of 
lines  dropped  from  points  in  A  1>  of  elevation  to  A  K. 

as  shown  by  the  figures  2  to  t>.      Likewise  set  off  1'r 

1,1  "ii  (^  I {  the  length  of  solid  lines  in  plan,  as  shown  l>y 
the  small  figures  7  to  12.  With  the  "["-square  parallel 
with  P  Q,  ereet  lines  from  the  points  in  0  1!.  and, 
measuring  from  ()  It.  set  oil'  on  these  lines  the  length 
of  lilies  of  corresponding  number  in  E  1?  V  of  elevation, 


/•'/;/.  ;.'.•'.  —  lii'ii/i-tnii  <ifs,  •:•!  ions  mi 
the  Solid  Linen  «f  the  I'lun. 


/•'/(/.  7411.—  liitttjmm  of  Sections  on 
the  Dotted  Lines  of  the  Plan. 


and  connect  the  points  thus  obtained  with  the  points 
in  P  Q,  as  indicated  by  the  numbers  in  the  plan.  Thus 
connect  <;  with  7.  ;">  with  8,  4  with  9,  etc. 

To  construct  a  diagram  of  sections  corresponding 
to  the  dotted  lines  of  the  plan  draw  the  right  angle 
ST  I'.  Fig.  740,  and  set  off  on  S  T  the  length  of  lines 
dropped  from  points  in  A  B  of  elevation  to  A  F.  or 
tranl'er  the  distances  in  P  Q,  Fig.  73!».  On  T  U  set 
off  the  length  of  dotted  lines  in  plan,  and  from  the 
points  thus  obtained  draw  lines  parallel  with  S  T, 
making  these  lines  of  the  same  length  as  those  dropped 
from  points  in  E  13  of  elevation  to  E  F.  Connect  the 
upper  ends  of  lines  8  to  12  with  points  in  S  T,  as  indi- 
cated by  the  dotted  lines  in  plan.  Thus  connect  2 
with  12,  3  with  11,  -t  with  10,  etc. 

The  next  step  is  to  obtain  the  distance  between 
points  in  A  I>  and  K  B  of  elevation  as  if 'measured  upon 
the  curved  surfaces  of  the  wall.  It  is  therefore  neces- 
sary to  develop  extended  sections  of  the  two  curves  of 
the  arch  as  shown  at  the  left  in  the  engraving.  The 
development  of  the  curve  of  the  inner  side  of  the  arch 
is  projected  directly  from  the  elevation  in  the  follow- 
ing manner:  ( )u  A  1''  extended  lav  off  J  K,  equal  to 
the  curve  1  line  A'  II  of  plan,  miking  the  straight  line 
equal  to  the  stretchout  of  half  of  the  curve  of  the  plan, 
as  indicated  by  the  small  figures.  At  right  angles  to 

J  O  CO 

J  K,   and  from  the  points  in  the  same,   erect  lines, 


making  them  the  same  length  as  lines  of  similar  num- 
ber dropped  from  points  in  A  B,  or,  with  the  "["-square 
parallel  with  J  K  and  A  F.  earrv  lines  from  the  points 
iu  A  B  intersecting  the  vertical  lines  of  similar  num- 
ber. A  line  traced  through  the  points  of  intersection, 
as  shown  by  .1  L.  will  be  the  desired  shape.  The  shape 
on  F'  K' of  the  plan,  corresponding  to  E  15  F  of  eleva- 
tion, is  obtained  in  a  similar  manner.  On  M  X  in  Fig. 
7:Js  set  off  the  stretchout  of  E'  F'  of  plan,  as  indicated 
by  the  small  ligures,  from  the  points  of  which  erect 
vertical  lines.  On  these  lines  set  off  the  same  length 
as  the  lines  of  similar  number  in  E  B  F  of  elevation. 
A  line  traced  through  the  points  thus  obtained  will 
give  the  desired  outer  curve  of  the  arch. 

To  develop  the  pattern  from  tne  several  sections 
obtained  proceed  in  the  following  manner:  Draw  the 
line  a  e  of  Fig.  7-H,  in  length  equal  to  Q  K  of  Fig. 
739,  and  with  e  as  center,  and  M  12  of  the  curve  M  O 
as  radius,  strike  a  small  arc,  12,  which  intersect  with 
one  struck  from  <t  as  center,  and  Q  12  of  Fig.  I'.W  as 
radius,  thus  establishing  the  point  12  of  pattern. 
With  n  of  pattern  as  center,  and  J  2  in  the  curve  J  L 
as  radius,  describe  a  small  arc.  2,  which  intersect  with 
one  struck  from  point  12  of  pattern  as  center,  and  212 
of  Fig.  740  as  radius,  thus  establishing  the  point  2  of 
pattern.  With  2  of  pattern  as  center,  and  2  11  of 
Fig.  739  as  radius,  strike  another  small  arc,  which 
intersect  with  one  struck  from  point  12  of  pattern  as 
center,  and  12  11  in  the  curve  M  O  as  radius,  thus 
establishing  the  point  11  of  pattern.  Continue  in  this 
way,  using  the  tops  of  the  sections  in  Figs.  739  and 


t\g.  741.— Half  Pattern  of  Soffit. 

740  for  measurements  across  the  pattern,  and  the 
spaces  in  the  inner  and  outer  curves  as  developed  in 
J  L  and  M  0,  Fig.  738,  for  the  distances  about  the 
edges  of  the  pattern,  establishing  the  several  points, 
as  shown,  through  which  draw  the  lines  e  f,  a  h  and 
f  /(,  thus  completing  the  pattern  for  one-half  the  soffit 
of  the  arch.  The  other  half  of  pattern  can  be  ob- 
tained by  the  same  method  or  by  duplication. 


418 


The  New  Metal    Worker  Pattern  Book. 


PROBLEM  218. 


Pattern  of  the  Blank  for  a  Curved  Molding  in  an  Arch  in  a  Circular  Wall. 


In  the  last  paragraph  of  Problem  215  it  is  stated 
that  the  demonstration  there  given  is  applicable  to  the 
blank  for  a  curved  molding  in  a  circular  wall.  There 
are  many  forms  of  arches  and  different  methods  of 
adapting  them  to  the  requirements  of  a  curved  wall. 
An  arch  may  be  semicircular  or  elliptical,  having 
either  the  long  or  the  short  diameter  of  the  ellipse  as 
its  width;  and  in  either  case  it  may  be  rampant, 
though  rampant  arches  are  seldom  used.  Any  form 
of  arch  may  be  so  constructed  that  its  moldings  project 
either  from  the  exterior  or  from  the  interior  surface 
of  a  curved  wall.  In  adapting  any  form  of  arch  to  a 
circular  wall  the  soffits  and  roof  strip,  or  portions  which 
appear  level  at  the  top  of  the  arch,  may,  as  they  are 
carried  around  the  curve  of  the  arch,  arrive  at  the 
springing  line  or  top  of  the  impost  parallel  to  the  cen- 
ter line  in  plan ;  or  they  may  at  this  point  be  drawn 
radially  to  the  curve  of  the  wall. 

In  Fig.  742  is  shown  one-half  of  the  elevation 
and  plan  of  a  semicircular  window  cap  and  a  section 
on  the  center  line  of  the  same.  M  K  L  N  is  one- 
half  the  elevation  of  all  the  members  constituting  the 
molding,  the  lines  of  which  are  projected  from  the 
profile  W  as  shown  to  the  right  of  the  center  line.  In 
the  plan,  those  lines  which  in  profile  W  were  drawn 
horizontally  are  here  drawn  parallel  to  the  center  line 
B  G.  They  might  with  equal  propriety  have  been 
drawn  radially  toward  the  center  of  the  curve  E  C. 
Should  they  be  drawn  radially  the  profile  of  the  mold- 
ing would  remain  nearly  normal  throughout  its  course, 
but  when  drawn  as  in  the  plan,  Fig.  742,  it  will  be  seen 
that  the  profile  is  continually  changing,  that  at  the 
foot  of  the  arch  or  top  of  the  corbel  being  shown  at  V 
in  the  plan.  This  profile  is  obtained  by  the  usual 
operation  of  raking,  as  shown  by  the  dotted  lines. 

As  the  blank  for  any  curved  molding  is,  to  the 
pattern  cutter,  a  flaring  strip  of  metal,  it  simply  be- 
comes necessary  to  determine  its  width  and  the  amount 
of  flare  which  it  may  assume  in  the  different  parts  of 
its  course,  after  which  its  pattern  may  be  arrived  at 
by  methods  described  in  Problem  214  and  those  fol- 
lowing. The  direction  of  the  line  determining  the 
amount  of  flare  necessary  for  the  strip  to  have  is  de- 
termined by  the  judgment  of  the  pattern  cutter  and 
the  requirements  of  the  machinery  used  in  "raising" 


the  mold.  Therefore  in  making  the  application  ot 
the  demonstration  given  in  Problem  -2\~>  to  the  win- 
dow cap  shown  in  Fig.  742  it  is  necessary  to  lirst 
draw  tin-  lines  /• .«/ of  the  profile  in  elevation,  ami  pr 
of  the  plan,  establishing  the  flare  and  necessary  width 
or  stretchout  at  those  points,  after  which  the  points 


M  8     76 

5 

t 
3 

1  N 

a 

fig.  742.—  Elevation  and  Plan  of  an  Arched  Wind**"-   <'HI<  '"    " 
Circular   Wall. 

ji  and  r  may  be  dropped  upon  the  line  M  N  of  the 
elevation,  and  the  points  k  and  //  earned  hori/.ontalk 
to  the  center  line  K  X,  as  shown.  The  points  thus 
obtained  in  M  N  and  K  L  must  then  be  connected 
bv  the  necessary  ares  struck  from  X  as  center,  thus 
completing  an  elevation  of  the  llaring  piece.  Likewise 
the  projection  of  the  points  //  and  /;  from  K  L  must 
l.e  sel  off  from  C  on  C  G  of  the  plan,  as  shown  at  >/" 
and  &",  and  the  points  thus  obtained  connected  with 


Pattern  Problems. 


419 


r  and  p  by  arcs  struck  from  the  center  used  in  describ- 
ing the  plan  of  tlie  wall. 

Should  the  width  and  llaiv  of  the  blank,  as  deter- 
mined by  the  raked  jirolile  Y  at  the  foot  of  the  aivh. 
vary  so  much  from  those  of  the  normal  jiroiile  at  the 
top  that  parallel  curved  lines  could  not  be  drawn  to 
connect  the  points  at  the  foot  with  those  at  the  top  in 
either  or  both  views,  then  centers  must  bo  found  upon 


73(>;  after  which  the  demonstration  given  in  that 
problem  may  be  followed  to  obtain  the  required 
pattern. 

It  will  thus  be  seen  from  the  foregoing  that  no 
matter  what  form  of  arch  be  used  or  in  what  manner 
it  be  placed  upon  the  wall,  the  method  of  obtaining 
the  pattern  of  its  flaring  surfaces  remains  the  same. 

It  might  under  some  circumstances  be  desirable 


Fig.  743.— Method  of  Obtaining  a  Section  through  Molding. 


the  center  lines  of  the  plan  and  of  the  elevation  from 
which  arcs  can  be  drawn  connecting  the  required 
points.  Having  thus  completed  a  plan  and  elevation  of 
the  flaring  piece  or  blank,  it  will-be  seen  by  reference 
to  Problem  215  that  the  lines  p'  r  and  k'  g'  of  the 
elevation,  Fig.  742,  correspond  with  A  ¥  and  B  E  of 
Fig.  730;  and  that  p  r  of  the  plan.  Fig.  742.  and  the 
ares  drawn  from  it  correspond  with  IV  A'.  V  K'  of  Fig. 


to  construct  stays  at  several  intervals  between  the  top 
and  the  foot  of  an  arch,  similar  to  that  shown  at  Y  in 
Fig.  742,  for  the  purpose  of  more  accurately  determin- 
ing the  flare  in  all  parts  of  its  sweep,or  for  the  purpose 
of  constructing  a  form  of  templet  to  assist  in  the 
operation  of  raising  the  mold.  The  profiles  of  such 
stays  can  be  obtained  by  the  usual  operation  of  rak- 
ing, which  is  fully  described  in  numerous  problems  in 


420 


Tlie  Kew  Metal   Worker  Pattern  Book. 


the  first  section  of   this  chapter.      In   Fig.    743    the 
operation  of  obtaining  a  profile  upon  the  line  X'  X"  is 


graving    must    be    obtained  by   developing  extended 
sections   of   those   lines    on  E  C   of   the   plan,    as   de- 


Fig.  744- — Perspective  View  nf  Templet  for   Use  in  Raising  the   Curved  Mold. 


carried    out    in    all    its    detail,   resulting  in   the  form 
shown  at  Q,  and  needs  no  further  comment. 

In   constructing  a   form   or  templet,  as   shown  in 
Fig.    74-4,  the   outlines  of   the  arch  shown  in  the  en- 


scribed  in  Problem  214  and  those  following,  after 
which  the  stays  can  be  placed  upon  lines  drawn  to 
correspond  with  those  from  which  the  respective  sec- 
tions were  taken. 


Note.— An  alphabetical  list  of  Terms,  arranged  as  an  index,  will  be  found  on  page  15.      The   Geometrical   Problems  giveu 
in  Chapter  IV  are  not  indexed,  but  as  each  problem  is  illustrated  by  a  special  diagram,  its  nature  may  be  readily  determined. 


Problems.   Page. 

ADJACENT  GABLES.— The  miter  between  the  mold- 
ings of  adjacent  gables  of  different  pitches  upon 
an  octagon  pinnacle 98  208 

ADJACENT  GABLES.— The  miter  between  the  mold- 
ings of  adjacent  gables  of  different  pitches  upon 
a  pinnacle  with  rectangular  shaft 97  207 

ARCH.— Pattern  for  the  soffit  of  an  arch  in  a  cir- 
cular wall,  the  soffit  being  level  at  the  top  and 
the  jambs  of  the  opening  being  splayed  on  the 
inside 217  416 

ARCH. —Pattern  for  a  splayed  elliptical  arch  in  a 
circular  wall,  the  opening  being  larger  on  the 
outside  of  the  wall  than  on  the  inside 215  412 

ARCH  —Pattern  for  a  splayed  arch  in  a  circular 
wall,  the  larger  opening  being  on  the  inside 
of  the  wall 216  414 

ANVIL.— The  patterns  for  an  anvil 63        162 

ARTICLE.— The  pattern  for  an  article  forming  a 
transition  from  a  rectangular  base  to  an  ellip- 
tical top 178  330 

ARTICLE.—  The  pattern  for  an  article  forming  a 
transition  from  a  rectangular  base  to  a  round 
top,  the  top  not  being  centrally  placed  over  the 
base 179  331 

ARTICLE  —The  pattern  for  an  article  rectangular 
at  one  end  and  round  at  the  other,  the  plane  of 
the  round  end  being  parallel  to  that  of  the  rec- 
tangular end 182  337 

ARTICLE.— The  pattern  for  an  article  with  rectan- 
gular base  and  round  top 177  328 

ARTICLE. — Pattern  of  an    article  having  elliptical 

base  and  round  top 185        343 

BALL. — The  pattern  for  the  miter  between  the 
moldings  of  adjacent  gables  upon  a  square  shaft 
formed  by  means  of  a  ball 73  176 

BALL.— To  construct  a  ball  in  any  number  of  pieces 

of  the  shape  of  gores 30        ]24 

BALL.— To  construct  a  ball  in  any  number  of  pieces 

of  the  shape  of  zones 124        351 

BATH.  —Pattern  for  a  hip  bath  of  regular  flare 143        373 

BATH  — The  patterns  for  a  hip  bath 172        313 

BATHTUB — The  patterns  for  a  bathtub 197       366 


Problems.  Page. 
BATHTUB.— Pattern  for  the  lining  of  the  head  of  a 

bathtub 303        378 

BIFURCATED  PIPE.— The  patterns  for  a  bifurcated 
pipe,  the  two  arms  being  the  same  diameter  as 
the  main  pipe  and  leaving  it  at  the  same  angle 45  137 

BLANK  for  a  curved  molding 126       254 

BLANK.— Pattern  for  the  blank  for  a  curved  mold- 
ing in  an  arch  in  a  circular  wall 218  418 

BLOWER  FOR  A  GRATE.— The  pattern  for  a  blower 

for  a  grate 73       180 

BOILER  COVER  —Pattern  for  a  raised  boiler  cover 

with  rounded  corners 165        309 

Boss.— The  pattern  for  a  conical  boss 141        271 

Boss.— The  pattern  for  a  boss  to  fit  around  a  faucet 203       379 

Boss.— The  pattern  of  a  boss  fitting  over  a  miter  in 

"a  molding 35       120 

BRACKET.— The  patterns  for  a  raking  bracket 88        193 

BROKEN  PEDIMENT.— To  obtain  the  profile  and  the 
patterns  of  the  returns  at  the  top  and  foot  of  a 
segmental  broken  pediment 94  204 

BROKEN  PEDIMENT.— To  obtain  the  profile  of  a 
horizontal  return  at  the  top  of  a  broken  pedi- 
ment necessary  to  miter  with  a  given  inclined 
molding  and  the  patterns  of  both 93  202 

BUTT  MITER  against  a  curved  surface 5         99 

BUTT  MITER  against  an  irregular  or  molded  surface 8       103 

BUTT  MITER  against  a  plain  surface  oblique  in  ele- 
vation    i  97 

BUTT  MITER  against  a  plain  surface  oblique  in  plan 2         97 

BUTT  MITER  of  a  molding  inclined  in   elevation 

against  a  plain  surface  oblique  in  plan 70        173 

CAN  Boss.— Pattern  for  a  can  boss  to  fit  around 

a  faucet 79       131 

CAPITAL.— The  construction  of  a  volute  for  a  capital 49       142 

CHIMNEY  TOP.— The  pattern  for  a  chimney  top 176       327 

CIRCULAR  WALL.— Pattern  for  a  splayed  arch  in  a 
circular  wall,  the  larger  opening  being  on  the 
inside  of  the  wall 216  414 

CIRCULAR  WALL.— Pattern  for  a  splayed  elliptical 
arch  in  a  circular  wall,  the  opening  being  larger 
on  the  outside  of  the  wall  than  on  the  inside 215  412 


422 


Index  of  Probknls. 


Problems.   Page. 

CIRCULAR   WALL. — Pattern    of    the   blank    for   a 
»    curved  molding  in  an  arch  in  a  circular  wall 218        418 

CIRCULAR  WALL.— Pattern  for  the  soffit  of  an  arch 
in  a  circular  wall,  the  soffit  being  level  at  the 
top  and  the  jambs  of  the  opening  being  splayed 
on  the  insi  !e 217  416 

CIRCULAR  WALL.— Pattern  for  the  soffit  of  a  semi- 
circular arch  in  a  circular  wall,  the  soffit  being 
level  at  the  top  and  the  jambs  of  the  opening 
being  at  right  angles  with  the  walls  in  plan. 
Two  cases 214  408 

COAL  HOD.— The  patterns  for  a  funnel  coal  hod ....  21 1        399 

COLD  AIR  Box. — The  patterns  for  a  cold  air  box  in 
which  the  inclined  portion  joins  the  level  por- 
tion obliquely  in  plan.  Two  solutions 99  210 

COLD  AIR  Box.— The  patterns  for  the  inclined  por- 
tion of  a  cold  air  box  to  meet  the  horizontal 
portion  obliquely  in  plan ...  100  214 

COLLAR. — The  pattern  for  a  collar,  round  at  the  top 
and  square  at  the  bottom,  to  fit  around  a  pipe 
passing  through  an  inclined  roof 180  333 

COLLAR. — The  pattern  for  a  flaring  collar,  the  top 
and  bottom  of  which  are  round  and  placed 
obliquely  to  each  other 194  361 

CONE. — Pattern  for  the  frustum  of  a  cone  fitting 

against  a  surface  of  two  inclinations 139        269 

CONK.— Patterns  of  a  cylinder  joining  a  cone  of 
greater  diameter  than  itself  at  right  angles  to 
the  side  of  the  cone 159  296 

CONES. — The  patterns  of  the  frustums  of  two  cones 

of  unequal  diameters  intersecting  obliquely 162        302 

CONES. — The  patterns  of  two  cones  of  unequal  di  • 
ameter  intersecting  at  right  angles  of  their 
axes 161  300 

CONE. — The  envelope  of  a  frustum  of  an  elliptical 

cone  having  an  irregular  base 174        322 

CONE. — The  envelope  of  a  frustum  of  a  right  cone 123       250 

CONE. — The  envelope  of  a  frustum  of  a  right  cone 

contained  between  planes  oblique  to  its  axis 144       274 

CONE.  —The  envelope  of  an  elliptical  cone 164       308 

CONE. — The  envelope  of  a  right  cone 122       249 

CONE. — The  envelope  of  a  right  cone  cut  by  a  plane 

parallel  to  its  ax's 147        278 

CONE. — The  envelope  of  a  right  cone  whose  base  is 

oblique  to  its  axis 136        265 

CONE.  — The  envelope  of  a  scalene  cone 163       306 

CONE. — The   envelope   of   the  frustum  of   a  cone 

the  base  of  which  is  an  elliptical  figure 132        260 

CONE. — The  envelope  of  the  frustum  of  a  right 
cone  the  upper  plane  of  which  is  oblique  to  its 
axis 135  265 

CONE. — The  envelope  of  the  frustum  of  a  scalene 167       311 

CONE. — The  frustum  of  a  cone  intersecting  a  cyl- 
inder of  greater  diameter  than  itself  at  other 
than  right  angles 155  290 


Problems.   Page. 
CONE. — The  frustum  of  a  cone  whose  base  is  a  true 

ellipse ...171        317 

CONE. — The  pattern  of  a  cone  intersected  by  a  cyl- 
inder at  its  upper  end,  their  axes  crossing  at 
right  angles 145  275 

CONE. — The  pattern  of  a  frustum  of  a  cone  inter- 
sected at  its  lower  end  by  a  cylinder,  their  axes 
intersecting  at  right  angles 140  270 

CONE. — The  pattern  of  a  tapering  article  with  equal 
flare  throughout  which  corresponds  to  the  frus- 
tum of  a  cone  whose  base  is  an  approximate 
ellipse  struck  from  centers,  the  upper  plane  of 
the  frustum  being  oblique  to  the  axis 1-16  276 

CONE. — The  patterns  of  a  cone  intersected  by  a  cyl- 
inder of  less  diameter  than  itself  at  right  angles 
to  its  base,  the  axis  of  the  cylinder  being  to  one 
side  of  that  of  the  cone 158  294 

COSE.— The  patterns  of  a  cone  intersected  by  a  cyl- 
inder of  less  diameter  than  itself,  their  axes 
crossing  at  right  angles 157  29.3 

CONE. — The  patterns  of  a  cylinder  joining  the  frus- 
tum of  a  cone  in  which  the  axis  of  the  cylinder 
is  neither  at  right  angles  to  the  axis  nor  to  the 
side  of  the  cone 160  298 

CONE.— The  patterns  of  the  frustum  of  a  cone  join- 
ing a  cylinder  of  greater  diameter  than  itself 
at  other  than  right  angles,  the  axis  of  the  frus- 
tum passing  to  one  side  of  that  of  the  cylinder 156  291 

CONE. — The  patterns  of  the  frustum  of  a  scalene 
cone  intersected  obliquely  by  a  cylinder,  their 
axes  not  lying  in  the  same  plane 175 

CONICAL  Boss.— The  pattern  for  a  conical  boss 141 

CONICAL  SPIRE. — The  pattern  of   a   conical    spire 

mitering  upon  eight  gables 151 

CONICAL  SPIRE.— The   pattern    of  a  conical   spire 

mitering  upon  four  gables 150 


CORNER  PIECE. — Patterns  for  the  corner  piece  of 
a  mansard  roof  embodying  the  principles  upon 
which  all  Mansard  finishes  are  developed 11 

COVER. — The  pattern  for  a  raised  boiler  cover  with 

rounded  corners 165 

COVER. — The  pattern  of  an  oblong  raised  cover  with 

semicircular  ends 129 

CRACKER  BOAT. — The  pattern  for  a  cracker  boat 138 

CURVED  MOLDINGS  —The  patterns  for  simple  curved 

moldings  in  a  window  cap 127 

CURVED  MOLDING. — Pattern    of   the  blank    for  ;\ 

curved  molding  in  an  arch  in  a  circular  wall 218 

CURVED  MOLDING. — The  blank  for  a  curved  mold- 
ing  126 

CURVED  MOLDING. — The  pattern   for   the   curved 

molding  in  an  elliptical  window  cap 128 

CURVES.— To  obtain  the  curves  for  a  molding  cov- 
ering the  hip  of  a  curved  mansard  roof Ill 

CYLINDER.— Patterns  of  a  cylinder  joining  a  cone  of 
greater  diameter  than  itself  at  right  angles  to 
the  side  of  the  cone 159 


323 

271 

282 
280 

10.V 
309 

253 
267 

265 

418 
25! 
256 
238 

296 


Index  of  Problems, 


423 


Problems.  Page. 

CYLINDER.— The  pattern  of  a  cone  intersected  by  a 
cylinder  at  its  upper  end,  their  axes  crossing  at 
right  angles 145  275 

CYLINDER.— The  patterns  of  a  cone  intersected  by  a 
cylinder  of  less  diameter  than  itself  at  right 
angles  to  its  base,  the  axis  of  the  cylinder  being 
to  one  side  of  that  of  the  cone 158  294 

CYLINDER.— The  patterns  of  a  cylinder  joining  the 
frustum  of  a  cone  in  which  the  axis  of  the  cyl- 
inder is  neither  at  right  angles  to  the  axis  nor 
to  the  side  of  the  cone 160  2P8 

CYLINDER.— The  patterns  of  a  cylinder  mitering  with 
the  peak  of  a  gable  coping  having  a  double 
wash 69  171 

CYLINDER. — The  patterns  of  the  frustum  of  a  scalene 
cone  intersected  obliquely  by  a  cylinder,  their 
axes  not  lying  in  the  same  plane 175  323 

DECAGON. — The  patterns  for  a  newel  post,  the  plan 

of  which  is  a  decagon 22        117 

DIAGONAL  BRACKET  —The  patterns  for  a  diagonal 

bracket  under  the  cornice  of  a  hipped  roof 90        198 

DODECAGON.— The  patterns  for  an  urn,  the  plan  of 

which  is  a  dodecagon 23        118 

DORMER. — Pattern  for  the  molding  on  the  side  of  a 
dormer,  mitering  against  the  octagonal  side  of  a 
tower  roof 65  165 

DOUBLE  ELBOW.— The  pattern  for  the  intermediate 
piece  of  a  double  elbow  joining  two  other 
pieces  not  lying  in  the  same  plane 53  147 

DROP. — The  pattern  for  a  drop  upon  the  face  of  a 

bracket .24        119 

ELBOW. — Patterns  for  a  right  angle  piece  elbow  to 

connect  a  round  with  a  rectangular  pipe 213        405 

ELBOW. — Patterns    for   a  three-piece    elbow    in    a 

tapering  pipe 153        285 

ELBOW. — Patterns  for  a  three-piece  elbow  to  join  a 

round  pipe  with  an  elliptical  pipe 212        403 

ELBOW. — Patterns  for  a  two-piece  elbow  in  a  taper- 
ing pipe 152  284 

ELBOW.— The  pattern  for  the  intermediate  piece  of 
a  double  elbow,  joining  two  other  pieces  not  lay- 
ing in  the  same  plane 53  147 

ELBOW. — The  patterns  for  a  five-piece  elbow 42        133 

ELBOW. — The  patterns  for  a  four-piece  elbow 41        133 

ELBOW. — The  patterns  for  an  elbow  at  any  angle 44        136 

ELBOW. — The  patterns  for  a  regular  tapering  elbow 

in  five  pieces 154        287 

ELBOW. — The  patterns  for  a  right  an'gle  two  piece 
elbow,  one  end  of  which  is  round  and  the  other 
elliptical 206  387 

ELBOW. — The  patterns  for  a  three-piece  elbow 40        131 

ELBOW. — The  patterns  for  a  three-piece  elbow,  the 

middle  piece  of  which  tapers 191        355 

ELBOW.— The  pattern  for  a  two-piece  elbow 38       130 


Problems.  Page. 
ELBOW. — The  patterns  for  a  two-piece  elbow  in  an 

elliptical  pipe.     Two  cases 39        130 

ELLIPTICAL  CONE. — The  envelope  of  a  frustum  of 

an  elliptical  cone  hiving  an  irregular  base 174        822 

ELLIPTICAL  CONE.— The  envelope  of  an  elliptical 

cone 164        308 

ELLIPTICAL  PIPE.— Joint  between  an  elliptical  pipe 
and  a  round  pipe  of  larger  diameter  at  other  than 
right  angles.  Two  cases 58  155 

ELLIPTICAL  PIPE.— The  pattern  of  an  elliptical  pipe 

to  fit  against  a  roof  of  one  inclination 32        125 

ELLIPTICAL  VASE.— The  patterns  for  an  elliptical 

vase  constructed  in  twelve  pieces 81        183 

EQUILATERAL  TRIANGLE.— A  pattern  for  a  pedestal 

of  which  the  plan  is  an  equilateral  triangle 16        111 

FACE  AND  SIDE.— Patterns  of  the  face  and  side  of  a 

plain  tapering  keystone 10        104 

FACE  MITER  at  right  angles,  as  in  the  molding 

around  a  panel 12        106 

FINIAL.— The  patterns  for  a  finial  the  plan  of  which 

is  an  irregular  polygon 82        185 

FINIAL.— The  patterns  of  a  finial  the  plan  of  which 

is  octagon  with  alternate  long  and  short  sides 85        189 

FIVE-PIECE  ELBOW.— The  patterns  for  a  five-piece 

elbow 43        133 

FLANGE. — A  conical  flange  to  fit  around  a  pipe  and 

against  a  roof  of  one  inclination 137        266 

FLANGE. — The  pattern  for  a  flaring  flange  round  at 
the  bottom,  the  top  to  fit  a  round  pipe  passing 
through  an  inclined  roof 195  363 

FLANGE.— The    pattern    for  a  flaring  flange  to  fit 
a  round  pipe  passing  through  an  inclined  roof 
.   the  flange  to  have  an  equal  projection  from  the 
pipe  on  all  sides 193        363 

FLANGE. — The  pattern  for  a  pyramidal  flange  to  fit 
against  the  sides  of  a  round  pipe  which  passes 
through  its  apex  50  144 

FLANGE. — The  pattern  of  a  flange  to  fit  around  a  pipe 

and  against  a  roof  of  one  inclination 37        129 

FLANGE. — The  pattern  of  a  flange  to  fit  around  a  pipe 

and  over  the  ridge  of  a  roof 36        128 

FLARING  ARTICLE. — Pattern  for  an  irregular  flaring 
article  which  is  elliptical  at  the  base  and  round 
at  the  top,  the  top  being  so  situated  as  to  be  tan- 
gent to  one  end  of  the  base  when  viewed  in  plan . . .  186  345 

FLARING  ARTICLE.— Pattern  for  an  irregular  flaring 
article  whose  top  is  a  circle  and  whose  base  is  a 
quadrant 190  354 

FLARING  ARTICLE. — Pattern  of  an  irregular  flaring 
article  both  top  and  bottom  of  which  are  round 
and  parallel  but  placed  eccentrically  in  plan. 
Otherwise  the  envelope  of  the  frustum  of  a  sca- 
lene cone 167  311 

FLARING  ARTICLE. — The  pattern  for  an  irregular 
flaring  article,  elliptical  at  the-  base  and  round 


424 


Index  of  Problems. 


Problems.  Page. 

at  the  top,  the  top  and  bottom  not  being  paral- 
lel  196        364 

FLARING  ARTICLE. — The  pattern  for  a  flaring  ar- 
ticle or  transition  piece,  round  at  the  top  and 
oblong  at  the  bottom,  the  two  ends  being  con- 
centric in  plan 187  346 

FLARING  ARTICLE. — The  pattern  for  a  flaring  article, 
round  at  the  top  and  bottom,  the  top  being 
placed  to  one  side  of  the  center  as  seen  in  plan 183  339 

FLARING  ARTICLE.— The  pattern  for  a  flaring  ar- 
ticle round  at  the  base  and  square  at  the  top 181  335 

FLARING  ARTICLE. — The  pattern  for  a  flaring  ar- 
ticle, round  at  top  and  bottom,  one  side  being 
vertical 184  341 

FLARING  ARTICLE. — The  pattern  of  a  flaring  article 

of  which  the  base  is  an  oblong  and  top  square 74        177 

FLARING  ARTICLE. — The  pattern  of  a  rectangular 

flaring  article 9        104 

FLARING  ARTICLE.  The  pattern  of  a  regular  flar- 
ing article  which  is  oblong  with  semicircular 
ends 130  258 

FLARING  ARTICLE. — The  pattern  of  a  flaring  article, 
the  top  of  which  is  round  and  the  bottom  ob- 
long, with  semicircular  end.  Two  cases 168  312 

FLARING  ARTICLE.  —The  pattern  of  a  flaring  article 
which  corresponds  to  the  frustum  of  a  cone 
whose  base  is  a  true  ellipse 171  317 

FLARING  ARTICLE. — The  pattern  of  a  flaring  article 
which  is  rectangular,  with  rounded  corners, 
having  more  flare  at  the  ends  than  at  the  side? 170  316 

FLARING  COLLAR  — The  pattern  for  a  flaring  collar, 
the  top  and  bottom  of  which  are  round  and 
placed  obliquely  to  each  other 194  361 

FLARING  END.— The  pattern  of  the  flaring  end  of  an 

oblong  tub 76        178 

FLARING  FLANGE.— The  pattern  for  a  flaring  flange, 
round  at  the  bottom,  the  top  to  fit  a  round  pipe 
passing  through  an  inclined  roof 195  363 

FLARING  FLANGE.— The  pattern  for  a  flaring  flange 
to  fit  a  round  pipe  passing  through  an  inclined 
roof,  the  flange  to  have  an  equal  projection 
from  the  pipe  on  all  sides 198  368 

FLARING  OBLONG  ARTICLE. — The  pattern  of  a  reg- 
ular flaring  oblong  article  with  round  corners 131  259 

FLARING  PAN. — The  pattern  of  an    oval   or   egg 

shaped  flaring  pan 134        263 

FLARING  SECTION.— The  pattern  of  the  flaring  sec- 
tion of  a  locomotive  boiler 77  179 

FLARING  SHAPE. — Pattern  for  an  irregular  flaring 
shape  forming  a  transition  from  a  round  hori- 
zontal base  to  a  round  top  placed  vertically 201  376 

FLARING  TRAY. — The   pattern    of   a   heart-shaped 

flaring  tray 133       261 

FLARING  TUB. — The  patterns  of  a  flaring  tub  with 
tapering  sides  and  semicircular  head,  the  head 
having  more  flare  than  the  sides 169  314 


Problems.  Page. 
FLOAT.— The  patterns  for  a  soap  makers'  float 173       320 

FORGE.— Pattern  for  the  hood  of  a  portable  forge 199        371 

FORK. — Pattern  for  a  three-prong  fork  with  taper- 
ing branches 2'  8  391 

FOUR-PIECE  ELBOW.— The  patterns  for  a  four-piece 

elbow 41        132 

FOUR- PIECE  ELBOW.- The  patterns  for  a  pipe  inter- 
secting a  four-piece  elbow  through  one  of  the 
miters 61  159 

FOUR-SIDED  FIGURE. — The  patterns  of  a  molding 

mitering  around  an  irregular  four-sided  figure 14        109 

FRUSTUM  OF  A  CONE  intersecting  a  cylinder  of  greater 

diameter  than  itself  at  other  than  right  angles 155        290 

FRUSTUM  OP  A  CONE.— The  envelope  of  the  frustum 

of  a  cone  the  base  of  which  is  an  elliptical  figure. .  .132        260 

FRUSTUM  OF  A  CONE  whose  base  is  a  true  ellipse 171       317 

FRUSTUM  OF  A  PYRAMID.— The  envelope  of  the  frus- 
tum of  a  pyramid  which  is  diamond  shape  in 
plan  75  178 

FRUSTUM  OF  A  RIGHT  CONE. — The  envelope  of  th<> 
frustum  of  a  right  cone  the  upper  plane  of  which 
is  oblique  to  its  axis 135  265 

FRUSTUM. — Pattern  of  a  tapering  article  with  equal 
flare  throughout  which  corresponds  to  the  frus- 
tum of  a  cone  whose  base  is  an  approximate  el- 
lipse struck  from  centers  the  upper  plane  of  the 
frustum  being  oblique  to  the  axis 146  276 

FRUSTUMS. — The  patterns  of  the  frustums  of  two 

cones  of  unequal  diameters  intersecting  obliquely. . .  162        302 

FRUSTUM. — The  envelope  of  a  frustum  of  an  ellip- 
tical cone  having  an  irregular  base 174  322 

FRUSTUM. — The  envelope  of  a  frustum  of  a  right 

cone 123        250 

FRUSTUM. — The  envelope  of  a  frustum  of  a  right 
cone  contained  between  planes  oblique  to  its 
axis 144  274 

FRUSTUM. — The  envelope  of  the  frustum  of  an  octag- 
onal pyramid 116  243 

FRUSTUM. — The  envelope  of  the  frustum  of  a  scalene 

cone , 167        311 

FRUSTUM. — The  envelope  of  the  frustum  of  a  square 

pyramid 1 15        242 

FRUSTUM. — The  pattern  for  the  frustum  of  a  cone 

fitting  against  a  surface  of  two  inclinations 139        269 

FRUSTUM. — The  pattern  of  a  frustum  of  a  cone  in- 
tersected at  its  lower  end  by  a  cylinder,  their 
axes  intersecting  at  right  angles 140  270 

FRUSTUM. — The  patterns  of  the  frustum  of  a  cone 
joining  a  cylinder  of  greater  diameter  than  itself 
at  other  than  right  angles  the  axis  of  the  frus- 
tum passing  to  one  side  of  that  of  the  cylinder 156  291 

FRUSTUM. — The  patterns  of  the  frustum  of  a  scalene 
cone  intersected  obliquely  by  a  cylinder,  their 
axes  not  lying  in  the  same  plane 175  323 


Index  of  Problems. 


425 


Problems. 
FUNNEL  COAL  HOD.— The  patterns  for  a  funnel  coal 

hod on 

GABLE  COPING. — The  patterns  for  a  square  shaft  of 
curved  profile,  mitering  over  the  peak  of  a 
gable  coping,  having  double  wash 68 

GABLE  COPING.— The  patterns  of  a  cylinder  miter- 
ing  with  the  peak  of  a  gable  coping  having  a 
double  wash 69 

GABLE  CORNICE..— The  pattern  for  a  gable  cornice 

mitering  upon  an  inclined  roof 64 

GABLE  MITERS.— The    patterns    of     simple    gable 

miters 15 

GABLE  MOLDING.— The  pattern  for  a  gable  molding 

mitering  against  a  molded  pilaster 62 

GORE  PIECE. — The  pattern  for  a  gore  piece  forming- 
a  transition  from  an  octagon  to  a  square,  as  at 
the  end  of  a  chamfer 86  191 


Page. 
399 

169 

171 
164 
110 
161 


GORE  PIECE. — The  pattern  for  a  gore  piece  in  a 
molded  article  forming  the  transition  from  a 
square  to  an  octagon 87  192 

GORES. — To    construct   a   ball  in  any  number  of 

pieces  of  the  shape  of  gores 30        134 

GORE. — The  pattern  for  a  three  piece  elbow,  the 

middle  piece  being  a  gore S3        1^7 

GRATE.  —The  pattern  for  a  blower  for  a  grate 78        ISO 

HEXAGONAL  PYRAMID.— The  envelope  of  a  hexag- 
onal pyramid H4  341 

HEXAGON.— The  pattern  for  a  pedestal  the  plan  of 

which  is  a  hexagon 19        114 

HEPTAGON.— The  pattern  for  a  vase  the  plan  of 

which  is  a  heptagon. ...   20        115 

HIP  BAR.  —Patterns  for  the  top  and  bottom  of  the 

hip  bar  in  a  skylight ]03        219 

HIP  BATH.  —Pattern  for  a  hip  bath  of  regular  flare. . . .  143        273 
HIP  BATH.— The  patterns  for  a  hip  bath 172        313 

HIP  FINISH.— The  pattern  for  a  hip  finish  in  a 
curved  mansard  roof,  the  plan  of  the  hip  being 
a  right  angle fi  IQI 

HIP  MOLDING. — Pattern  for  a  hip  molding  mitering 
against  the  planceer  of  a  deck  cornice  on  a  man- 
sard roof  which  is  square  at  the  eaves  and  octa- 
gon at  the  top 107  227 

HIP  MOLDING. — The  pattern  for  a  hip  molding  upon 
a  right  angle  in  a  mansard  roof  mitering  against 
a  b°d  molding  at  the  top 102  217 

HIP  MOLDING. — Patterns  for  the  fasdas  of  a  hip 
molding  finishing  a  curved  mansard  roof  which 
is  square  at  the  base  and  octagonal  at  the  top no  236 

HIP  MOLDING. — Patterns  for  a  hip  molding  miteriug 
against  the  bed  molding  of  a  deck  cornice  on 
a  mansard  roof  which  is  square  at  the  base  and 
octagonal  at  the  top 108 


HIP  MOLDING. — The  pattern  of  a  hip  molding  upon 
an  octagon  angle  of  a  mansard  roof  mitering 
upon  an  inclined  wash  at  the  bottom 106 


230 


225 


Problems.    Page. 

HIP  MOLDING.— The  pattern  of  a  hip  molding  upon 
a  right  angle  in  a  mansard  roof  mitering  against, 
the  planceer  of  a  deck  cornice 101  215 

HIP  MOLDING.— The  patterns  for  the  miter  at  the 
bottom  of  a  hip  molding  on  a  mansard  roof 
which  is  octagon  at  the  top  and  square  at  the 
bottom 109  233 

HIP  MOLD.— The  pattern  of  a  hip  mold  upon  an  oc- 
tagon angle  in  a  mansard  roof  mitering  against 
a  bed  molding  of  corresponding  profile 105  223 

HOOD.— Pattern  for  the  hood  of  a  portable  forge 199        371 

HOOD.— The  patterns  for  the  hood  of  an  oil  tank 200        374 

HORIZONTAL  MOLDING.— From  the  profile  of  a  given 
horizontal  molding  to  obtain  the  profile  of  an  in- 
clined molding  necessary  to  miter  with  it  at  an 
octagon  angle  in  plan,  and  the  patterns  for  both 
arms  of  the  miter 95  205 

HORIZONTAL  MOLDING.— From  the  profile  of  a  given 
inclined  molding  to  establish  the  profile  of  a 
horizontal  molding  to  miter  with  it  at  an  octa- 
gon angle  in  plan,  and  the  patterns  for  both  arms. . ;  .96  206 

HORIZONTAL  RETURN.— To  obtain  the  profile  of  a 
horizontal  return  at  the  foot  of  a  gable  necessary 
to  initer  at  right  angles  in  plan  witli  an  inclined 
molding  of  normal  profile,  and  the  miter  patterns 
of  both 91  200 

HORIZONTAL  RETURN.— To  obtain  the  profile  of  a 
horizontal  return  at  the  top  of  a  broken  pedi- 
ment necessary  to  miter  with  a  given  inclined 
molding,  and  the  patterns  of  both 93  202 

INCLINED  MOLDING.— From  the  profile  of  a  given 
horizontal  molding  to  obtain  the  profile  of  an  in- 
clined molding  necessary  to  miter  with  it  at  an 
octagon  angle  in  plan,  and  the  patterns  for  both 
arms  of  the  miter 95  305 , 


INCLINED  MOLDING.—  The  pattern  for  an  inclined 
molding  mitering  upon  a  wash  including  a  re- 
turn .............................................  66  166 

INCLINED  MOLDLNG.--TO  obtain  the  profile  of  an  in- 
clined molding  necessary  to  miter  at  right  an 
gles  in  plan  with  a  given  horizontal  return,  and 
the  miter  patterns  of  both  ........................  93  OQI 


INCLINED  ROOF.—  The  pattern  for  a  gable  cornice 

mitering  upon  an  inclined  roof  ....................  64        164 

JACK  BAR.  —Pattern  for  the  top  of  a  jack  bar  in  a 

skylight  .........................................  104        221 

JOINT  at  other  than  right  angles  between  two  pipes 
of  different  diameters,  the  axis  of  the  smaller 
pipe  being  placed  to  one  side  of  that  of  the 
larger  one  .........................................  60  158 

JOINT  between  an  elliptical  pipe  and  a  round  pipe  of 
larger  diameter  at  other  than  right  angles. 
Two  cases  ........................................  53  155 

JOINT  between  two  pipes  of  different  diameters  in- 

tersecting at  other  than  right  angles  ...............  57        154 

JOINT  between  two  pipes  of  the  same  diameter  at 

other  than  right  angles  ............................  54       IQI 


426 


Index  of  Prollefns. 


Problems.    Page. 
JOINT.— The  patterns  for  a  T- joint  between  pipes  of 

the  same  diameter 47        139 

JUNCTION. — Patterns  for  the  junction  of  a  large 
pipe  with  the  elbows  of  two  smaller  pipes  of  the 
same  diameter 205  384 

KEYSTONE. — The  patterns  for  a  keystone  having  a 

molded  face  with  sink 26        120 

LEVEL  MOLDING. — The  pattern  for  a  level  molding 
mitering  obliquely  against  another  level  mold- 
ing of  different  profile 67  167 

LIP.— Pattern  for  the  lip  of  a  sheet  metal  pitcher 142        272 

LOCOMOTIVE  BOILER.— The  pattern  for  the  flaring 

section  of  a  locomotive  boiler 77        179 

MANSARD  ROOF. — Pattern  for  a  hip  molding  miter- 
ing  against  the  planceer  of  a  deck  cornice  on  a 
mansard  roof  which  is  square  at  eaves  and  oc- 
tagon at  the  top 107  227 

MANSARD  ROOF. — Patterns  for  a  hip  molding  miter- 
ing  against  the  bed  molding  of  a  deck  cornice 
on  a  mansard  roof  which  is  square  at  the  base 
and  octagonal  at  the  top 108  230 

MANSARD  ROOF. — Patterns  for  the  corner  piece  of  a 
inaneard  roof  embodying  the  principles  upon 
which  all  mansard  finishes  are  developed 11  105 

MANSARD  ROOF.— The  pattern  for  a  hip  finish  in  a 
curved  mansard  roof  the  plan  of  the  hip  being 
aright  angle 6  101 

MANSARD  ROOF. — The  pattern  for  a  hip  molding 
upon  a  right  angle  in  a  mansard  roof  mitering 
against  a  bed  molding  at  the  top ;102  217 

MANSARD  ROOF. — The  pattern  of  a  hip  molding  upon 
an  octagon  angle  of  a  mansard  roof  mitering 
upon  an  inclined  wash  at  the  bottom 106  225 

MANSARD  ROOF. — The  pattern  of  a  hip  molding  upon 
a  right  angle  in. a  mansard  roof  mitering  against 
the  planceer  of  a  deck  cornice 101  215 

MANSARD  ROOF. — The  pattern  of  a  hip  mold  upon 
an  octagon  angle  in  a  mansard  roof  mitering 
against  a  bed  molding  of  corresponding  profile 105  223 

MANSARD  ROOF. — The  patterns  for  the  fascias  of  a 
hip  molding  finishing  a  curved  mansard  roof 
which  is  square  at  the  base  and  octagonal  at  the 


top. 


MANSARD  ROOF. — The  patterns  for  the  miter  at  the 
bottom  of  a  hip  molding  on  a  mansard  roof 
which  is  octagon  at  the  top  and  square  at  the 
bottom 109 

MANSARD  ROOF.— To  obtain  the  curves  for  a  mold- 
ing covering  the  hips  of  a  curved  mansard  roof. ...  Ill 

MITER. — A  butt  miter  against  a  curved  surface 5 

MITER. — A  butt  miter  against  an  irregular  or  molded 
surface 


.110        236 


MITER.— A  butt  miter  against  a  plain  surface  oblique 
in  elevation 


8 


.1 


MITER. — A  butt  miter  against  a  plain  surface  oblique 
in  plan 


233 

238 
99 

103 
97 
97 


Problems.    Page. 

MITER.— A  face  miter  at  right  angles,  as  in  the  mold- 
ing around  a  panel 12  106 

MITER. — A  return  miter  at  other  than  a  right  angle, 

as  in  a  cornice  at  the  corner  of  a  building 4          98 

MITER.— A  square  return  miter  or  a  miter  at  right 
angles,  as  in  a  cornice  at  the  corner  of  a  build- 
ing  3  98 

MITER  between  two  moldings  of  different  profiles 7        102 

MITER  LINE. — To  obtain  the  miter  line  and  pattern 
for  a  straight  molding  meeting  a  curved  mold- 
ing of  the  same  profile 55  153 

MITERS. — Patterns  of  simple  gable  miters 15        110 

r.IiTER.— The  pattern  for  the  miter  between  the 
moldings  of  adjacent  gables  upon  a  square  shaft 
formed  by  means  of  a  ball 73  176 

MOLDED  BASE. — The  patterns  for  a  molded  base  in 
which  the  projection  of  the  sides  is  different 
from  that  of  the  ends 80  182 

MOLDED  FACE. — The  patterns  for  a  keystone  having 

a  molded  face 26        120 

MOLDED  PILASTER. — The  pattern  f  or  a  gable  molding 

mitering  against  a  molded  pilaster fi2        161 

MOLDING. — A  butt  miter  of  a  molding  inclined  in  el- 
evation against  a  plain  surface  oblique  in  plan 70  173 

MOLDINGS.— Miter  between  two  moldings  of  differ- 
ent profiles 7  102 

MOLDING. — Pattern  for  the  molding  on  the  side  of 
a  dormer  mitering  against  the  octagonal  side  of 
a  tower  roof 65  165 

MOLDINGS. — Pattern    for    the    moldings    and    roof 

pieces  in  the  gables  of  an  octagon  pinnacle' 72        174 

MOLDINGS. — Patterns    for    the    moldings    and  roof 

pieces  in  the  gables  of  a  square  pinnacle ...  71        173 

MOLDING. — The  patterns    of    a    molding    mitering 

around  an  irregular  four-sided  figure 14        109 

MOLDINGS. — The  patterns  of  the  moldings  bound- 
ing a  panel  triangular  in  shape 13  108 

MOLDING. — To  obtain  the  niiter  line  and  pattern  for 
a  straight  molding  meeting  a  curved  molding  of 
the  same  profile 55  15'3 

NEWEL  POST. — The  patterns  for  a  newel  post  the 

plan  of  which  is  a  decagon 22        117 

OBLONG  TUB. — The  pattern  of  the  flaring  end  of  an 

oblong  tub 76        178 

OBLONG  VESSEL. — Pattern  for  the  end  of  an  oblong 
vessel  which  is  semicircular  at  the  top  and  rec- 
tangular at  the  bottom 166  310 

OCTAGONAL  PEDESTAL. — The  patterns  for  an  octag- 
onal pedestal 21  115 

OCTAGONAL  PYRAMID. — Envelope  of  the  frustum  of 
an  octagonal  pyramid  having  alternate  long 
and  short  sides 117  244 

OCTAGONAL  PYRAMID. — The  mvelopeof  the  frustum 

of  an  octagonal  pyramid 116        243 


Index  of  Problems, 


427 


Problems. 

OCTAGONAL  SHAFT.— The  patterns  of  an  octagonal 
shaft  the  profile  of  which  is  curved  fitting  over 
the  ridge  of  a  roof 29 

OCTAGON  PINNACLE.  —Pattern  for  the  moldings  and 

roof  pieces  in  the  gables  of  an  octagon  pinnacle 72 

OCTAGON  PINNACLE.— The  miter  between  the  mold- 
ings of  adjacent  gables  upon  an  octagon  pinna- 
cle   98 

OCTAGON  SHAFT  mitering  upon  the  ridge  and  hips  of 

a  roof 35 

OCTAGON  SHAFT.— The  pattern  of  an  octagon  shaft 

fitting  over  the  ridge  of  a  roof 33 

OCTAGON  SHAFT. — To  describe  the  pattern  of  an  oc- 
tagon shaft  to«fit  against  a  ball 28 

OCTAGON  SPIRE. — The  pattern  for  an  octagon  spire 
mitering  upon  a  roof  at  the  junction  of  the 
ridge  and  hips 121 

OCTAGON  SPIRE. — The  pattern  of  an  octagon  spire 

mitering  upon  eight  gables 119 

OCTAGON  SPIRE. — The  pattern  of  an  octagon  spire 

mitering  upon  four  gables 120 

OCTAGON.. — The  patterns  of  a  finial  the  plan  of  which 

is  octagon  with  alternate  long  and  short  sides 85 

OFFSET. — Pattern  for  an  offset  to  join  a  round  pipe 

with  one  of  elliptical  profile 210 

OFFSET. — The  pattern  for  an  offset  between  two 
pipes,  oblong  in  section,  whose  long  diameters 
meet  at  right  angles  to  each  other 1 89 

OFFSET. — The  pattern  for  an  offset  to  join  an  oblong 

pipe  with  a  round  one 209 

PAX. — The  pattern  of  an  oval  or  egg-shaped  flaring 

pan 134 

PANEL. — Patterns  of  the  moldings  bounding  a  panel 

triangular  in  shape 13 

PEDESTAL.  — The  pattern   for   a  pedestal  of  which 

the  plan  is  an  equilateral  triangle 16 

PEDESTAL. — The  pattern  for  a  pedestal   square  in 

plan 17 

PEDESTAL. — The  pattern  for  a  pedestal  the  plan  of 

which  is  a  hexagon 19 

PEDESTAL.— The  patterns  for  an  octagonal  pedestal 21 

PENTAGON.— The  patterns  for  a  vase  the  plan  of 

which  is  a  pentagon 18 

PINNACLE.  -^  The  miter  between  the  moldings  of  ad- 
jacent gables  of  different  pitches  upon  a  pin- 
nacle wHli  rectangular  shaft 97 

pIPEs._ A  joint  between  two  pipes  of  the  same  di- 
ameter at  other  than  right  angles 54 

PIPES. — A  T-joint  between  pipes  of  different  diam- 
eters  i 56 

PIPES. — A  T-joint  between  pipes  of  different  diam- 
eters, the  axis  of  the  smaller  pipe  passing  to  one 
side  of  that  of  the  larger 59 


Page. 

128 

174 

208 
127 
126 
122 

248 
245 
246 
189 
396 

350 
393 
263 

108 
111 
113 

114 
115 

112 

207 
151 
153 

156 


Problems. 

PIPES. — Joint  at  other  than  right  angles  between 
two  pipes  of  different  diameters,  the  axis  of  the 
smaller  pipe  being  placed  to  one  side  of  that  of 
the  larger  one 60 

PIPES.— The  joint  between  two  pipes  of  different 
diameters  intersecting  at  other  than  right  an- 
gles  57 

PIPE. — The  pattern  of  an  elliptical  pipe  to  fit  against 

a  roof  of  one  inclination 32 

PIPE. — The  pattern  of  a  flange  to  fit  around  a  pipe 

and  against  a  roof  of  one  inclination 37 

PIPE. — The  pattern  of  a  flange  to  fit  around  a  pipe 

and  over  the  ridge  of  a  roof 36 


PIPE. — The  pattern  of  a  round  pipe  to  fit  against  a 

roof  of  one  inclination 31 


158 

154 
125 
129 

128 
125 
127 

137 
134 
159 
146 
253 

140 

272 

185 


PIPE. — The  pattern  of  a  round  pipe  to  fit  over  the 

ridge  of  a  roof 34 

PIPE. — The  patterns  for  a  bifurcated  pipe,  the  two 
arms  being  of  the  same  diameter  as  the  main 
pipe  and  leaving  at  same  angle 45 

PIPE.— The  patterns  for  a  pipe  carried  arountt  a 

semicircle  by  means  of  cross  joints 43 

PIPE. — The  patterns  for  a  pipe  intersecting  a  four- 
piece  elbow  through  one  of  the  miters 61 

PIPE.  —The  patterns  for  a  rectangular  pipe  intersect- 
ing a  cylinder  obliquely 52 

PIPE. — The  patterns  for  a  semicircular  pipe  with 

longitudinal  seams 125 

PIPE. — The  patterns  for  a  square  pipe  describing  a 

twist  or  compound  curve 48 

PITCHER.  — Pattern  for  the  lip  of  a  sheet  metal  pitcher ...  1 42 

POLYGON.— The  patterns  for  a  finial  the  plan  of 

which  is  an  irregular  polygon 82 

PROFILE. — From  the  profile  of  a  given  horizontal 
molding  to  obtain  the  profile  of  an  inclined  mold- 
ing necessary  to  miter  with  it  at  an  octagonal 
angle  in  plan  and  the  patterns  for  both  arms  of 
the  miter 95  205 

PROFILE. — From  the  profile  of  a  given  inclined  mold- 
ing to  establish  the  profile  of  a  horizontal  mold- 
ing to  miter  with  it  at  an  octagon  angle  in  plan 
and  the  patterns  for  both  arms 96  206 

PROFILE.— To  obtain  the  profile  and  patterns  of  the 
returns  at  the  top  and  foot  of  a  segmental 
broken  pediment 94  204 

PROFILE. — To  obtain  the  profile  of  a  horizontal  re- 
turn at  the  foot  of  a  gable  necessary  to  miter  at 
right  angles  in  plan  with  an  inclined  molding 
of  normal  profile,  and  the  miter  patterns  of 
both 91  200 

PROFILE. — To  obtain  the  profile  of  a  horizontal  re- 
turn at  the  top  of  a  broken  pediment  necessary 
to  miter  with  a  given  inclined  molding,  and  the 
patterns  of  both . . ." 93  202 

PROFILE. — To  obtain  the  profile  of  an  inclined  mold- 
ing necessary  to  miter  at  right  angles  in  plan 


428 


ProbUms. 

with  a  given  horizontal  return,  and  the  miter 
patterns  of  both 93 

PYRAMIDAL  FLANGE.— The  pattern  for  a  pyramidal 
flange  to  fit  against  the  sides  of  a  round  pipe 
which  passes  through  its  apex r>"i 

PYRAMID.—  The  envelope  of  the  frustum  of  a  square 

pyramid 1 15 

PYRAMID.  —The  envelope  of  a  hexagonal  pyramid 1 14 

PYRAMID  — The  envelope  of  a  square  pyramid 1 1 3 

PYRAMID.— The  envelope  of  a  triangular  pyramid 1:2 

PYRAMID.— The  envelope  of  the  frustum  of  an  octag- 
onal pyramid 1  K; 

PYRAMID.— The  envelope  of  the  frustum  of  an  oc- 
tagonal pyramid  having  alternate  long  and 
short  sides 117 

PYRAMID.— The  evelope  of  the  frustum  of  a  pyra- 
mid which  is  diamond  shape  in  plan 7.3 

PYRAMID.— The  patterns  for  a  square  pyramid  to 
fit  against  the  sides  of  an  elliptical  pipe  which 
passes  through  its  center 51 

RAKING    BRACKET. —The    patterns    for    a    raking 

bracket  88 

RAKING     BRACKET.— The    patterns    for    a     raking 

bracket  in  a  curved  pediment  192 

RAISED  PANEL  —The  pattern  for  a  raised  panel  on 

the  face  of  a  raking  brack-t 89 

RECTANGULAR  FLARINO  ARTICLE. -The  pattern  of 

a  rectangular  flaring  article. ...    9 

RECTANGULAR  PIPE.  -The  patterns  for  a  rectan- 
gular pipe  intersecting  a  cylinder  obliquely 53 

RETURN  MITER  at  other  than  a  right  angle,  as  i  i  a 

cornice  at  the  corner  of  a  building I 

RETURNS.  —To  obtain  the  profile  and  patterns  of  the 
returns  at  the  top  and  foot  of  a  segmental 
broken  pediment. 9j 

RIGHT  CONE.— The  envelope  of  a  right  co:io 123 

RIGHT  CONE.— The  envelope  of  a  right  cone  cut  by 

a  plane  parallel  to  its  axis 147 

RIGHT   CONE.— The   envelope   of  a    frustum  of  a 

right  cone .03 

RIGHT  CONE.— The  envelope  of  a  frustum  of  a  right 
cone  contained  between  planes  oblique  to  its 

144 


of  Problems. 


Page. 
201 

141 

242 
241 
241 
240 

24:! 

241 
178 


RIGHT  CONE.— The  envelope  of  a  right  cone  whose 

base  is  oblique  to  its  axis 136 

RIGHT  CONE.— The  envelope  of  the  frustum  of  a 
right  cone  the  upper- plane  of  which  is  oblique 
to  its  axis 135 

ROUND  PIPE.— The  pattern  of  a  round  pipe  to  Ft 

against  a  roof  of  one  inclination   31 

ROUND  PIPE.— The  pattern  of  a  round  pipe   to  fit 

over  the  ridge  of  a  roof 34 

SCALENE  CONE.— The  envelope  of  a  scalene  cone 163 


274 


265 


125 

127 
306 


lit:! 
358 
190 
104 
140 
98 

204 
219 


Problems.   Page. 
SCALENE  CONE  —The  envelope  of  the  frustum  of  a 

scalene  cone 1(57        311 

SCALENE  CONE.— The  patterns  of  the  frustum  of  a 
scalene  cone  intersected  obliquely  by  a  cylinder, 
their  axes  not  lying  in  the  same  plane 175  323 

SCALE  SCOOP. — The  pattern  for  a  scale  scoop  having 

both  ends  alike 143        373 

SCALE  SCOOP.—  The  patterns  for  a  scale  sci,()p,  one 

end  of  which  is  funnel  shaped 149        379 

SEGME.NTAL  BROKEN  PEDIMENT. — To  obtain  the  pro- 
file and  patterns  of  the  returns  at  the  top  and 
foot  of  a  segmental  broken  pediment 94  304 

SHIP  VENTILATOR.— Patterns  for  a  ship  ventilator 

having  a  round  base  and  an  elliptical  mouth 20 1        381 

SIDE  —Patterns  of  the  face  and  side  of  a  plain  taper- 
ing keystone 10  104 

SKYLIGH  r  BAR.— The  patterns  for  the  top  and  bot- 
tom of  a  "  common  "  skylight  bar 46  138 

SKYLIGHT. -Pattern  for  the  top  of  a  jack  bar  in  a 

skylight 11,4        221 

SKYLIGHT.— The  patterns  for  the  top  and  bottom  of 

the  hip  bar  in  a  skylight 103        219 

SOAP  MAKKRS'   FLOAT.— The   patlerns   for  a   soap 

makers'  float 173        330 

SOFFIT. — Pattern  for  the  soffit  of  an  arch  in  a  circu- 
lar wall,  the  soffit  being  level  at  the  top  and  the 
jambs  of  the  opening  being  splayed  on  the  inside. .  .217  416 

SOFFIT.— Pattern  for  the  soffit  of  a  semicircular  arch 
in  a  circular  wall,  the  soffit  being  level  at  the 
top  and  the  jambs  of  the  opening  being  at  right 
angles  to  the  walls  in  plan.  T\\>>  cases 

SPLAYED  ARCH.— Pattern  for  a  splayed  arch  in  a 
circular  wall,  the  larger  opening  being  on  the 
inside  of  the  wall 315 

SPLAYED  ELLIPTICAL  ARCH.  —  P.  it  tern  for  a  splayed 
elliptical  arch  in  a  circular  wall,  the  opening  be- 
ing larger  on  the  outside  of  the  wall  tlian  on  the 
inside 315 

SPIRE. — The  pattern  for  an  octagon  spire  initering 
upon  a  roof  at  the  junction  of  the  ridge  and 
hips 121 

HPIRE.— The   pattern    of    a  conical   spire   miterin.,' 

upon  four  gables 150 


SPIRE. — The  pattern  of  a  conical  spire  initering  upon 

eight  gables ir,i 

SPIRE. — The  pattein  of  an   octagon  spire  initering 
upon  eight  gables 


119 

SPIRE. —The  pattern  of  an   octagon  spire  initering 

upon  four  gables 130 

SPIRE  — The   pattern   of   a   1-qiia.re    spire    initering 

upon  four  gables .118 

SQUARE  PINNACLE.— Patterns  Tor  the  moldings  and 

roof  pieces  in  the  gables  of  a  square  pinnacle 71 

SQUARE  PIPE. — The  patterns  for  a  square  pipe  de- 
scribing a  twist  or  compound  curve 49 


214        408 
414 

412 

946 

280 
282 
245 
246 
244 


173 


140 


In'!'. i-  i if 


429 


Problems.   Page. 

SQUARE  PYRAMID. — The  envelope  of  a  square  pyra- 
mid  113  241 

SQUAKE  PYRAMID.— The  envelope  of  the  frustum  of 

a  square  pyramid 115        242 

SQUARE  PYRAMID. — The  patterns  for  a  square  pyra- 
mid to  fit  against  the  sides  of  an  elliptical  pipe 
which  passes  through  its  center 51  145 

SQUARE  RETURN  MITER,  or  miter  at  right  angles,  as 

in  a  cornice  at  the  corner  of  a  building 3          98 

SQUARE  SHAFT. — The  patterns  for  a  square  shaft  of 
curved  profile  mitering  over  the  peak  of  a  gable 
coping  having  a  double  wash 68  169 

SQUARE  SHAFT. — The  pattern  of  a  square  shaft  to  fit 

against  a  sphere 27        122 

SQUARE  SPIRE. — The  pattern  of  a  square  spin'  miter- 
ing  upon  four  gables 118  244 

TASK.— The  patterns  for  the  hood  of  an  oil  tank 200        374 

TAPERING  ARTICLE.—  The  patterns  of  a  tapering  arti- 
cle which  is  square  at  the  base  and  octagonal  at 
the  top 84  188 

TAPERING  ARTICLE.— The  pattern  of  a  tapering  arti- 
cle with  equal  flare  throughout  which  corre- 
sponds to  the  frustum  of  a  cone  whose  base  is  an 
approximate  ellipse  struck  from  centers,  the 
upper  plane  of  the  frustum  being  oblique  to  the 
axis 146  276 

TAPERING  ELBOW. — The  patterns  for  a  regular  ta- 
pering elbow  in  five  pieces 154  287 

THREE-PIECE    ELBOW. — Patterns    for  a   three  piece 

elbow  in  a  tapering  pipe 153        285 

THREE-PIECE  ELBOW. --Patterns  for  a  three-piece 
elbow  to  join  a  round  pii>e  with  an  elliptical 
pipe 212  403 

THREE  PIECE  ELBOW. — The  pattern  fora  three-piece 

elbow,  the  middle  piece  being  a  gore 83        187 

THREE-PIECE  ELBOW. — The  patterns  for  a  three- 
piece  elbow 40  131 

THREE-PIECE  ELBOW. — The  patterns  for  a  three- 
piece  elbow  the  middle  piece  of  which  tapers 191  355 

THREE-PRONG  FORK. — Pattern  for  a  three-prong  fork 

with  tapering  branches 208        391 

T-Joint  between  pipes  of  different  diamaters 56        153 

T-Joiut  between  pipes  of  different  diameters,  the 
axis  of  the  smaller  pipe  passing  to  one  side  of 
that  of  the  larger 59  156 

T- Joint.  —The  patterns  for  a  T-joint  between  pipes  of 

the  same  diameters 47        139 

TRANSITION  from  a  round  horizontal  base  to  a  round 

top  placed  vertically 201        376 

TRANSITION  PIECE. — Pattern  for  a  transition  piece 
round  at  the  top  and  oblong  at  the  bottom.  Two 
cases 188  348 


Problems.   Page. 

TRANSITION  PIECE.— Pattern  for  a  transition  piece 
to  join  two  rou:i  1  pipes  of  unequal  dianr-terat 
an  angle 193  359 

TRANSITION  PIECE. — The  pattern  for  a  flaring  article 
or  transition  piece  round  at  the  top  and  oblong 
at  the  bottom,  the  two  ends  being  concentric  in 
plan 187  346 

TRANSITION.— The  pattern  for  a  gore  piece  forming 
a  transition  from  an  octagon  to  a  square,  as  at 
the  end  of  a  chamfer 86        191 

TRANSITION.— The  pattern  for  a  gore  piece  in  a 
molded  article  forming  a  transition  from  a 
square  to  an  octagon 87  192 

TRANSITION.— The  pattern  for  an  article  forming  a 
transition  from  a  rectangular  base  to  an  ellipti-' 
cal  top 178  330 

TRANSITION  —The  pattern  for  an  article  forming  a 
transition  from  a  rectangular  base  to  a  round 
top.  the  top  not  being  centrally  placed  over  the 
base 179  331 

TRAY.— The  pattern  of  a  heart-shaped  flaring  tray 138        261 

TRIANGULAR  PYRAMID.— The  envelope  of  a  trian- 
gular pyramid 112  240 

TWO-PIECE  ELBOW.— Patterns  for  a  two-piece  el- 
bow in  a  tapering  pipe 152  384 

Two  PIECE  ELBOW.— The  pattern  for  a  two  piece 

elbow 38        130 

TWO-PIECE  ELBOW.— The  patterns  for  a  two-piece 

elbow  in  an  elliptical  pipe.     Two  cases 39        130 

TWO-PIECE  ELBOW.— The  patterns  for  a  right  angle 
two  piece  elbow  one  end  of  which  is  round  and 
the  other  elliptical 206  387 

URN. — The  patterns  for  an  urn  the  plan  of  which  is 

a  dodecagon 23        118 

VASE. — The  pattern  for  a  vase  the  plan  of  which 

is  a  heptagon 20        115 

VASE.— The  pattern  for  a  vase  the  plan  of  which  is 

a  pentagon 18        112 

VENTILATOR. — Patterns  for  a  ship  ventilator  having 

a  round  base  and  an  elliptical  mouth 204        881 

VOLUTE.  —  The    construction   of   a    volute    for   a 

capital 49        142 

WASH. — The  pattern  for  an  inclined  molding  miter- 
ing  upon  a  wash,  including  a  return 66  166 

WINDOW  CAP. — The  pattern  for  the  curved  molding 

in  an  elliptical  window  cap 128        256 

WINDOW    CAP. —The  patterns    for  simple   curved 

moldings  in  a  window  cap 127        255 

Y. — The  patterns  for  a  Y  consisting  of  two  tapering 

pipes  joining  a  larger  pipe  at  an  angle 207        389 

ZONES. — To    construct    a  ball  in   any  number   of 

pieces  of  the  shape  of  zones 124        251 


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